If fluid pressure is result from collision then why we need depth? First we learn that the deeper you go into the water the more weight you feel from above, which then the weight translate by pressure.
And we also know that the water force is come from the collision of water molecule to the surface of the object. 
Why we find pressure of fluid from the total weight of water above us rather than the total collision-causing-force of water at the surface of an object ?
 A: "First we learn that the deeper you go into the water the more weight you feel from above." We know that this can't be the full story because 'pressure acts equally in all directions'. What this means is that the water pushes at right angles with the same force per unit area on any small element of surface at a given depth, no matter at what angle the surface is orientated. For example the water will push upwards on the horizontal bottom of a submerged brick with exactly the same force per unit area that it pushes downwards on the top of a submerged brick whose top is at the same depth as the bottom of the first brick!
How can this be? Simply put, the weight of the water column above squashes the water below very slightly, that is reduces the separation of the molecules slightly below their normal equilibrium separations, causing repulsive forces between them. These are responsible for the pressure that the liquid exerts on surfaces with which it is in contact. I'm not sure whether or not these forces would correctly be described as collision forces. 
In the case of a gas, as in the atmosphere, squashed as it is by its own weight, the force it exerts on any surface most certainly is caused by its molecules colliding with the surface. But in this case the squashing does not consist of forcing molecules closer than their equilibium separations - they remain, on average, many times further apart from that - but in ensuring that there are a huge number of molecules in motion per $\text{m}^3$ of space, and therefore many collision per second per $\text{m}^2$ with any surface.
A: When forces act over distances they do work, which changes the energy of an object. A pressure in particular does work over a change in volume rather than distance, which causes it to be a force-per-unit-area. It's the same idea, just when we talk about fluids we have to spread things like "mass" and "energy change" over larger amounts of space.
The question is, what amount of energy do you need to put into, for example, some water in order to expand a piston a little bit? Remember this question as we have to take one mental excursion before we answer it.
You are operating under the assumption that the fluid is very easily compressible so that you just need to deal with some sort of "local volume" that gets compressed, not the effects on the whole fluid. There may indeed be certain fluids which behave this way in the short term, although I do not know what they are off the top of my head. If I had to guess, I would guess that you are looking for a fluid with an especially low speed of sound, so that your expansions and compressions locally cannot be reflected globally over the timescales that you're interested in. But over long timescales (which may be fractions of a second or less) the fluid will respond and form a very similar equilibrium to whatever it has now. 
In fact for most of the more familiar fluids including water and air, you will find that they are approximately incompressible at the scales that you are used to. I recommend testing this by closing a thin plastic bottle that you don't care much about, and trying to compress it to, say, half of its volume: you will find that with air, compressing it is possible but requires a tremendous effort, enough that you can't get more than 10-20% compression with the force in your hands -- possibly with your entire weight you might be getting 50% compression. Filling the same bottle with water you will find that compression is no longer possible no matter how hard you squeeze it; you simply don't have the strength. So pressures which you generally do not create with your body on the air are needed to substantially compress air and you simply cannot create the pressures needed to substantially compress water.
If the fluid is approximately incompressible, what happens? Recall that we're talking about an energy difference. If you expand a piston underwater by a small change in its height $\delta h$ then you have to push a small volume of water $\delta V = A~\delta h$ out of the way. But if the water is incompressible then that water has to push other water out of the way, and that water has to push other water out of the way, and so on, until eventually some water is pushed all the way up to the surface where it has to push some air out of the way, and eventually that air needs to push other air out of the way, until we get to the top of the atmosphere. Let's ignore the air for a moment. Fortunately the water is approximately homogeneous and this energy for the water is $\delta E=(\rho_\text{water}~\delta V)~g~d$ where $d$ is your depth: the energy cost to push the water pushing the water is exactly the same as the energy cost to teleport the volume of water that you're displacing to the very surface. Again, this happens because water is so incompressible that you would have to move extremely fast and hard in order to create a small pocket of "doubly dense" water that would not, in the "short term", be able to shove water up to the top. Those pressure waves travel 1.5 km per second in water, to give a rough idea of the speeds you'd need.
Dividing through we find that the pressure must be $\delta E /\delta V = \rho_\text{water}~g~d.$ When we deal with the air we have approximately the same drop off for small elevations, the formula is still $p_0 - \rho_\text{air}~g~h,$ but if you go up a kilometer the air will be only maybe 90% as dense and the temperature might be 5°C cooler and such, so we generally either just use $p_0 = 1 \text{ atm}$ if that's a suitable pressure or else we use atmosphere tables if we are ascending/descending by many hundreds of meters.
