$\def\vg{\vec g} \def \vF{\vec F} \def \vR{\vec R}$
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.
