Why exactly is the pressure an object experiences at a given depth in a fluid? I am currently trying to work through why an object with smaller density than the fluid it is in floats upwards.
I know that the object experiences pressure in all directions because it has moved water. However already here i start to get confused:
Why doesn't the newly moved water simply increase the height of water in the water container? Is it because to do so, the moved water would have to move all the water molecules above it for that to happen, and therefore often decides to not do so? Does that too explain why the pressure gets larger the deeper we get? because more water molecules have to be moved to increase the height of the water?
Now I'm learning about this from KhanAcademy, and when they had a cube placed in water at given depth $h$, they say that the pressure it experiences is:
$$\frac {h\cdot\text{density of water}\cdot g}{\text{area of cube}}$$
Basically, the weight of the water above the cube, divided by the cube area.
But when they've previously said that the pressure is NOT caused by the amount of water above our object, this confuses me greatly. And while it DOES make sense for the top of our cube, since the pressure is $F/a$, and the force it experiences IS the weight of the water above it, the same formula is also used to find the pressure at the bottom of the cube, and on the sides of the cubes. And in that case it makes no sense.
So why do we find the pressure with this formula when it seems contradictory with the reason our object experiences pressure?
 A: Consider a very small cube of water of side length $\Delta h$, density $\rho$ with the gravitational field strength $g$ in equilibrium with surrounding water.
The weight of that cube of water is $\Delta h^3\rho g$.  
As the cube of water is in equilibrium then the net force on it must be zero.  
Consider vertical forces.
On the top face there is a force $f$ downwards, the weight of the cube is downwards and there is a force $F$ on the bottom face of the cube upwards.
Using Newton's second law with up as positive gives $F - \Delta h^3\rho g - f=0 \Rightarrow \dfrac {F}{\Delta h^2} - \dfrac{f}{\Delta h^2} = \Delta h \rho g$.
Define a quantity called pressure as the force per unit normal area.  
$\dfrac {F}{\Delta h^2} - \dfrac{f}{\Delta h^2}$ is the difference in pressure across the cube $\Delta P = \Delta h\rho g$
That formula would still be the same if the cube of water was surrounded by a vessel made out of glass with a square base of side $\Delta h$ with an open top and above the water there was a vacuum - ignore the practicalities of doing this.
The vacuum would exert no force on the top of the cube of water and the base of the glass vessel would exert an upward force on the the base of the cube of water equal to the weight of the cube.
The pressure difference between the top and bottom of the cube of water would still be $\Delta P = \Delta h \rho g$.  
You can then extend this to lots of cubes of water piled on top of one another.
A: It makes perfect sense that pressure is still increasing with height at the bottom and sides of the cube.
Pressure acts equally in all directions, not just up/down.  It also is reasonable to consider it goes up as you have more water above you.  This even applies when the body isn't fully submerged, because the pressure still acts on the bottom and sides.  That pressure still changes with height.
The only reason pressure varies with height is because gravity is acting on it, so there is more than just internal forces of the liquid causing pressure.  If there were no potential energy associated with height (i.e. somewhere with simulated or real free-fall) then pressure shouldn't vary with height.
I'm not sure if this answers your question though, it's somewhat unclear to me what was confusing you.
A: Assuming a ideal (non viscous and non compressible),
The pressure $P$ at a depth $h$ of fluid with density $\rho$ on a planet with gravitational acceleration $g$ and atmospheric pressure $P_0$ assuming the upper surface (open end) of fluid is kept open to its atmosphere is
$$P=P_0 + \rho gh$$
JMac's answer covers the rest of your questions.
