Why Pascal's Law is true and what is the mechanism for force amplification at molecular level? 
"A change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid" 
"A pressure change occurring anywhere in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere"


I'm having a hard time getting the intuition behind Pascal's Law. I've already read many kinds of explanations, involving energy, work and displacement and read almost all the posts on this subject on Physics.SE but I can't convince myself. 
The main problems for me are to understand the force amplification as a consequence of Pascal's Law and the reason why the pressure is transmitted undiminished. I made a drawing to illustrate what's going on in my mind (where the balls represent atoms):




Assumptions in this post:


*

*The fluid is incompressible and non viscous;

*The container is perfectly rigid;

*There are no displacements involved (I emphasize this assumption because many of the explanations I read end up using the hydraulic press and ending with the argument "You have to pay for the multiplied output force by exerting the smaller input force through a larger distance", like these ones on Physics.SE: Force amplification and Newton's third law and What is the explanation of Pascal's law at molecular level?).



So, in a more explicit way, Pascal's Law says WHAT happen, but I want to know WHY it happen:


*

*In terms of forces, at molecular level, how pressure is transmitted undiminished to all points in the fluid, causing the amplification of the force when area increases?


$$\bbox[5px,border:1px solid black] {F_{top}=100\,N\xrightarrow{\,\uparrow{A}\,} F_{bottom}=1000\,N}$$


*

*In terms of energy, how does the energy associated with the force applied at the top spreads and turns into the force applied at the bottom? (Note: I know that energy, given my assumptions, is conserved | Note 2: just to remember - there are no displacements involved)


$$\bbox[5px,border:1px solid black] {100\,N=F_{top}\rightarrow E_{top}=E_{bottom}\rightarrow F_{bottom}=1000\,N}$$
 A: $\def\vg{\vec g} \def \vF{\vec F} \def \vR{\vec R}$
Warning: This answer does not take into account the long series of
comments preceding it. Therefore some repetition will be unavoidable.
My aim is to give - I hope - an orderly treatment of the matter.
External forces and stress
Your figures are mistaken for two reasons. First, you draw many small
arrows directed downward, as if pressure were a vector. It isn't.
Pressure has no direction.
Second, your first figure applies to a solid, not to a fluid (liquid
or gas). For solids Pascal's law doesn't hold, nor you may simply speak
of pressure: the useful concept is stress tensor, a much more
complicated thing.
Let's see however how reasoning would work in the solid case. You have
a block of solid matter, in form of a truncated cone. Suppose it is
resting on a table. Then resultant force on it must vanish. Which are
the forces acting on the block? The first is its weight, vertical
downward, resultant $m\vg$. The second is table's "reaction" $\vR$,
actually a distributed force, applied all over the contact surface.
Since net force is vanishing, we know that $\vR=-m\vg$. No forces
exist on the oblique lateral face.
In a solid atoms cannot move freely. They may only effect small
displacements from their equilibrium positions: thermal vibrations and
average net displacements if external forces are acting as in our
case. That the latter exist means that no solid is absolutely rigid.
Such displacements are necessary to produce forces between atoms, which
globally counteract external forces.
Let me explain better this delicate point. In static conditions not only
the resultant of external forces must vanish, but the same must happen
for whichever portion of the body. If you mentally isolate that
portion, you will see it to be subjected to two kinds of forces:


*

*force distributed all over the portion's volume (typically it is weight)

*surface force, acting between atoms residing at the two sides of the
portion's boundary.
It is the second kind of forces which at a macroscopic level are
represented by the stress tensor. They are generally not normal to the
boundary's surface, nor have the same intensity and direction from
one point to another of the body.

A solid block
Suppose now you apply a force $\vF$, vertically downward, to the upper
base of block. It is easy to understand that at equilibrium $\vR$ will
change, becoming 
$$\vR = -m \vg - \vF.$$ 
Note that this is true independently of where exactly the force is
applied. You may also split it in several sub-forces applied in
different points. Or you may think of a force continuously distributed
over the entire base. In the latter case it is customary to speak of
an applied "pressure" and this the reason why pressure is often
thought of (wrongly) as a vector.
What can be said about $\vR$? I mean, $\vR$ is the resultant of the
forces the table applies to block's lower base. But how are these
forces distributed? The question cannot be answered with the given
data. In mechanics jargon, this is a "statically undetermined
problem".
Caution: something more could be said by keeping into account the
moments of applied forces. However my last statement remains true.
Of course since we do not know the exact distribution of forces in the
lower base, the same happens for internal stresses in the block, both
when $\vF=0$ and when $\vF\ne0$. This shows the intrinsic complexity
of our problem when a solid body is concerned. 

Fluid and pressure
And now let's come back (finally! you will say) to the case of a
fluid. There are two main differences between this case and that of a
solid. First, a fluid needs to be contained in a vessel (a closed one
for a gas). Second, what defines a fluid for our problem is that
surface forces can only be orthogonal to the surface.
The first difference entails a complication: we have to take into
account forces applied by the vessel's lateral wall to the fluid. And
if these are not given, it could seem that the problem becomes
insoluble.
The second difference instead amounts to a great simplification,
because internal stress can be wholly described by a single scalar:
pressure. Let's see why is it so, from a microscopic point of view. A
solid can transmit a shear force, i.e. a force parallel to the
surface where it is applied. The reason is that atoms are only allowed
small displacements around their equilibrium positions, but there is
no constraint as to the displacement's direction. Therefore atoms near
boundary between two portions of the body (see above) may well displace
parallel to the surface and in opposite directions on opposite sides
of it.
On the contrary atoms (or molecules) in a fluid are more or less free
to move around, not being constrained in the vicinity of some point.
As a consequence the fluid's bulk cannot resist to a shear stress: it
immediately gives in and cancels the stress. 
Then in a fluid only normal stresses are allowed. But there is more:
it can be shown that when this situation prevails, then in a given
point the intensity of (normal) force is always the same, whichever
direction you choose for the boundary between two portions of the
body. Shortly, we say that stresses are isotropic. Thus we have
arrived at the pressure concept.
Pressure is not a force, has no direction (is a scalar). It does not
act at the surface of a body, but is present in every internal
point. In fact, it is well known that we may have pressure in an
unbounded gas. The omnipresent instance is our atmosphere, but think of
stars too: they are gigantic gas masses with no bounds, held together
only thanks their own gravity.

The case of gases
It is well to open a short parenthesis to mark an important difference
between liquids and gases. In liquids atoms are very near to each
other, and what we see as macroscopic forces may be correctly
interpreted as the resultant of microscopical forces between them.
Not so for gases, where distances are much higher, so that macroscopic
forces are better seen as the effect of a myriad of collisions in
which atoms exchange momentum one with another or with the vessel's
walls.

No rôle for viscosity
Another clarification is in order. I spoke of shearing stresses and of
their absence in fluids. Someone could think this is not so for a
viscous fluid. After all, viscosity is just defined as the ability of
real fluids to transmit shear stresses!
The answer is that we strictly bounded ourselves to static
situations. Viscosity only acts when a fluid is moving; it is a force
arising because parts of one fluid flow one wrt to another. The
classic example is a river, whose water runs faster at centre, whereas
grows slower nearing the bank, where it's still.
Therefore in static problems there is no need to restrict to
non-viscous fluids. Viscosity has no effect.

To begin with, let's neglect gravity
To understand Pascal's law it's well to neglect gravity, at least
initially. This could appear disconcerting, as there is a common
misconception that pressure is due to gravity. Many people believe that
atmospheric pressure is due to the weight of air above us (which in a
sense is true) and conclude (erroneously) that pressure acts "from
above". I'm afraid that not all introductory physics books are free
from such sin.
Curiously enough, those people forget that they are continuously using
objects which are counterexamples to that idea. I'm alluding to tires:
of cars, bicycles, and so on. All these are inflated with a pump
pressing air within. No rôle is played by atmosphere's weight.
A more exotic example is given by ISS, where a pressure is maintained
to keep astronauts' breathing comfortably. Yet there is almost no air
outside!
It is easy to show that in such situations pressure is the same in the
whole volume of your block, irrespective of the force you can apply to
movable parts of the vessel. "Wait a moment!" - I feel like I'm
hearing - "Force? Which force? Why should be a force?
It's better to begin with gases, easier to understand, I believe. If
the vessel is rigid no action is required by the experimenter.
Everything stands still, nothing happens. But we know (I said it
before) that the gas molecules continually hit the walls and rebound,
giving them some momentum. More exactly, a definite amount of momentum
per unit time and surface area. Momentum per unit time equals force.
Force per unit area equals pressure. Therefore this momentum exchange
is a measure of the gas pressure.
You may not notice that the wall is subjected to that force since
usually, if vessel is sufficiently rigid, it automatically develops
internal forces that counterbalance those due to gas and keep walls to
move or deform. But sometimes things go differently: a balloon
inflated at too high a pressure may blow. A welding in a metal tank
may leak...
In other cases the vessel is built with a movable part to make
experiments (the famous cylinder-with-piston of thermodynamics). This
case is obvious: the piston is steady only if force due to gas
pressure (force = pressure x area) is contrasted by an equal and
opposite force applied from outside.
For a liquid too things go in an analogous way. Instead of collisions
exchanging momentum we have forces between neighbouring molecules.
Those near a wall interact with piston's molecules and directly apply
forces to them. Result is the same: to keep piston steady an opposite
external force is needed. The bigger the piston's area, the stronger
the force.
If there are two pistons, the same argument applies to both, and we
easily conclude that the force required to keep a piston still is
proportional to its area. This is what was improperly called "force
amplification".

Increasing pressure
We can also see things the other way around. Gas or liquid pressure is
determined by the forces applied to piston(s). If you increase
external force(s) the fluid will momentarily give in. If liquid,
molecules will slightly get nearer one to another; this will increase
the repulsive forces between, until a new equilibrium is reached. For
a gas a volume reduction will result in a greater number of molecules
per unit volume, thus augmenting the number of collisions per unit
time against piston, i.e. augmenting pressure. Again, compression will
halt when equilibrium is attained.

Gravity comes into play
We may not always neglect gravity. Not for atmosphere only: scuba
divers know very well that underwater pressure increases by one
atmosphere every ten meters of depth. This contradicts what I said
before about pressure being the same at all points in a (still) fluid.
The increment of pressure is too easily attributed to weight of the
water column above. Sometimes this works, other times doesn't: see
hydrostatic paradox in the internet.
Actually what can be experimentally verified is the following law
(Stevin): in a fluid at equilibrium in a uniform gravitational field
$\vg$ the pressure difference between two any points is
$$p_1 - p_2 = \rho\,g\,(z_1 - z_2)$$
if $z$-axis is oriented like $\vg$. Just a simple example: if fluid is
water ($\rho=10^3\,\mathrm{kg/m^3}$) and $z_1-z_2=10\,\mathrm m\,$
then $p_1-p_2=98\,\mathrm{kPa}$ which is about 1 atm.
Of course Stevin's law is a consequence of fluid equilibrium under
internal and external forces already discussed, with gravity added. I
can't dwell on the proof, however.

What about energy?
I cant't close this extra long post without answering the above
question. It would be off topic if I'd used really static arguments,
but this is not so, since in several places I spoke of
"displacements". And when something is displaced with a force applied
to it, work is involved and therefore energy.
Here again a distinction must be made between gases and liquids,
because of their very different compressibilities. Under ordinary
pressures liquids may be assumed incompressible without significant
error. This is far from true for gases.
Of course, even an incompressible liquid can accomplish important
displacements. Incompressibility only means that overall volume does
not change. If the vessel has two pistons of different surface areas
it's easy to see that proportionality of force to area, derived above,
together with invariable volume, entail that works of external forces
on pistons are equal, save for sign. So total work done is zero and
liquid's energy does not change.
We could have reasoned in the converse: since energy must be
conserved, total work by external forces must be zero, then force is
proportional to area. But this argument has a flaw: work is done not
only by external forces. Internal forces too can do work. So we have
to prove that work of internal forces vanishes. This is not too easy,
and requires an inquiry at microscopic level. It is better to assume
it as a characteristic property of an incompressible fluid: no work is
required to displace it in any way (until kinetic energy is
negligible).
Note: I hope reader did notice that my energy argument rested on a
hypothesis: equality of pressures on both pistons. But we have seen
that this is not true in presence of gravity, if pistons are located
at different heights. Let me set aside this complication for now.
As to gases, constant volume cannot be assumed. Nothing forbids to
compress or dilate a gas. Furthermore, for gases temperature too
becomes important in this respect. But you are not expecting from me a
treatise on mechanics and thermodynamics of fluids ... did you?
When gas volume changes, work is done on it by external forces. The
relevant formula is well known: $W=-p\,\Delta V$. As I wrote it, this
formula is not generally true: it holds if $p$ stays constant during
displacement. Otherwise we should write an integral:
$$W = -\int_A^B \!\!p\,dV.$$
And this too requires that during transformation $p$, even if not
constant, is well defined and the same all over gas volume. Also
remember that if volume varies and work is done, to keep $p$ constant
energy must be drawn from or given to gas as heat flowing through walls.
Nothing more about gases. A short comment on what happens if pistons
are placed at different heights in a vessel containing a heavy fluid.
In this case pressures are different, and proportionality between force
and area is not respected. Then if pistons are moved, although
liquid's volume stays constant, work is done. A positive work if lower
piston is moved into liquid, higher piston in the opposite direction. 
Question: positive work means liquid gained energy. Where is it to be
found? Easy answer: liquid was generally lifted. More precisely, its
c.o.m. was lifted. Then liquid's potential energy in the gravity field has
increased. For a simple geometry, e.g. a parallelepiped vessel,
it is easy to prove that work equates increment of P.E. For a general
shape this holds still true, but the proof is more involved.
To conclude. We may say that energy plays no relevant rôle in
relation to Pascal's law. There is no "pressure energy" in a liquid,
although you will easily find such expression about Bernoulli's
theorem (which is out of my actual aim). Just to unravel the mystery:
what is improperly called pressure energy is enthalpy density. 
A: Fundamentally, you seem think of force as conserved stuff. I put 100N of force in, so I should get 100N of force out. The answer is to simply stop thinking of force that way. It's not stuff. It isn't conserved.
Suppose you have a long lever

The 5 kg mass lifts the 100 kg mass. But that's impossible! Where did all that extra force come from! Tell me at the molecular level! And none of this garbage about "displacement"!
Yeah, you could try to make up some overly-complicated story that fits those requirements. But a much better explanation is that if the 5kg mass falls 1cm, the 100 kg mass rises only 1/20 cm, so the total gravitational energy is unchanged. Energy behaves like stuff and an "amount in equals amount out" sort of rule applies. Force doesn't work that way.
For an analogy, imagine someone sells apples for \$1 in the market. In their house, they have some chairs that cost \$20 each. Would you demand that they explain how anyone could conceivably buy something that costs \$20 when they only get paid \$1 for what they sell? Would you demand they explain it at the level of how each penny moves? Of course not, because it's not a mystery and not hard to understand. What must be balanced is not the money per unit good, but the total money they take in. (money they make per apple)*(number of apples sold) > (money to buy a chair)*(number of chairs bought).
Force is like money per apple or money per chair. When we multiply force by displacement, we get energy, and that's conserved. If you see someone creating a large force somewhere, whether with an ordinary lever or with hydraulics, it doesn't make any more sense to demand an explanation of how they could possibly create such a large force than it does to demand an explanation of how someone could ever generate a large sum of money. They can generate a large force from a small force as long as they pay for it with a large displacement of the small force, the same as they can pay for an expensive chair with a large volume of apples sold.
But your post demands that no one is allowed to talk about displacement. Sorry, but that's wrongheaded. Displacement is a useful and important concept here. Your post further demands to know about the force at the molecular level, while stipulating that the fluid is incompressible. This just doesn't make sense. The pressure at the molecular level is explained by the repulsion of molecules from each other, which depends on how far apart they are. You can't have "incompressible" and "explain from the molecular level" at the same time.
The reason you were getting nowhere in your reading of previous answers isn't that those answers were bad. It's that you were making strong assumptions going into them - assumptions about what sort of thing force is - and those assumptions were wrong. It will be much more productive to revise your idea of force than to demand that everyone else's explanation kowtow to it.
A: Your graphic is not showing the contribution from the vessel itself.  It might be clearer to see that the angled walls can be approximated by a lot of small stairsteps.  Every little step sideways is accompanied by the vessel wall supplying pressure downward.
Or you can imagine a section of a wall supplying pressure inward, and decompose that pressure into sideways and downward components.  Either way is the same.  The walls will be found above 9m^2 of the bottom, and are pushing down on the fluid with 900N of force.

But that doesn't explain why the force is amplified.

Let me see if I can add some things so that it doesn't seem so strange.
First of all, we are only considering this to be a static system.  Because of this, you shouldn't think of the piston as a special source of force here.  It's not.  Every wall is pushing with the same pressure.  The force on the piston isn't causing the entire force on bottom surface, instead all of the surfaces are pushing at the same time. 
In fact, maybe you could imagine a vessel with 10 pistons on top.  9 are locked in place, and you push down on one.  All the other pistons will have the same pressure (and if the same size, the same force).  The only difference is that for the other pistons, the force is coming from the strength of the lock and the attachment to the vessel, not from your hand pushing down.  
In this case and in your case, there's 1000N pushing up on the fluid from below and 1000N pushing down on the fluid from above.  So there's no amplification of force.
We can change the question to: "why doesn't it matter how small the piston is"?
In this sense it does.  In order to get the fluid inside to a pressure of 100Pa, you had to compress it and do work.  (In the case of water or a substance that has very low compressibility, the work is very small, but is not zero).  The smaller the piston, the further you have to push it to get the same work and the same compression.  If the piston were tiny, you would need less force, but more distance and the amount of work done would be the same.
A: 
In terms of forces, at molecular level, how pressure is transmitted undiminished to all points in the fluid, causing the amplification of the force when area increases?

Remember your stipulation that there are no displacements. Then, for there to be no displacement within the fluid each drop of fluid must experience a balanced force. That means the force on the left must be equal and opposite to the force on the right, and so forth*. Since the cross sectional area is the same on the left and on the right, that implies that the pressure is also the same. Therefore the pressure is the same throughout the fluid. Since the pressure is undiminished this leads directly to the usual hydraulic force multiplication. 
*neglecting the weight of the fluid

In terms of energy, how does the energy associated with the force applied at the top spreads and turns into the force applied at the bottom? (Note: I know that energy, given my assumptions, is conserved | Note 2: just to remember - there are no displacements involved)

With no displacements there is no work and the energy is not relevant. You need to relax the no displacements stipulation to have a meaningful answer about energy. 
A: Pressure is omnidirecitonal because it is caused by the tranmission of energy between large numbers of tiny particles and the boundary under pressure. 
When momentum is imparted from one tiny particle to the other the direction changes, based on the specifics of the interaction (like billiard balls on a pool table) 
Each interaction is governed by conservation of momentum $m_1v_1+m_2v_2=m^{'}_1v^{'}_1+m^{'}_2v^{'}_2$
This, occuring randomly over large numbers of interactions, distributes the momentum (and therefore the kinetic energy) evenly in all directions.
Hence Pascal's Principle, pressure is the equal throughout the medium.
A: @Vinicius ACP wrote: 

if I am injecting 300N of pressure into the system, how the reaction forces of the walls can be more than 300N?

First of all, you didn't inject a pressure, you applied a force. Even
units are different. You applied that force on a portion of vessel's
walls. So doing, you forced liquid's molecules to get closer one to
another, in the whole volume of liquid. This is unavoidable, as I
already explained, since molecules can reach a new equilibrium position
only when each feels (in the average) a net zero force from its
neighbours. 
Incidentally, this doesn't happen instantaneously: a compression wave propagates within liquid and when rest is attained liquid's volume is slightly reduced. If you could measure liquid's density in various points, you would see that it's increased, although very little, of the same amount everywhere. 
Just to give you a concrete figure: if liquid is water, by increasing its pressure by $900\,\mathrm{Pa}$ you cause a relative density increment of about
$4\cdot10^{-7}$. 
Now molecules all around near vessel's walls are pushed by other molecules, but on the wall's side there are no other liquid molecules to contrast this push. In order to get equilibrium, push must be acted by the wall's molecules. And just to obey Newton's third law, those liquid's molecules must press the wall with an opposite force: $p\,dS$ on every surface element $dS$ of wall. 
Bottom included, of course.
