How does hydraulic press work at molecular level? Let's consider the following arrangement:
 
I know that because of conservation of energy the force on the car has to be amplified. The work $W$ done is
$$W = \text{Force} \cdot \text{distance} \, ,$$
which in turn is the amount of energy transferred.
How is this force amplified at the molecular level. For example, if we put sand in place of water and suppose that the molecules are connected with springs then can we show that the springs will get compressed in such a way so as to amplify the force on the car?
 A: from the figure: liquid that cannot be compressed... so forget compression of springs.
Molecularly, particles in a liquid have much less space to move around (short mean free paths) compared to particles in a gas. The result is that any applied pressure to these particles is instantaneously transferred through the fluid. The mechanism by which this happens is the collision of particles which exchange kinetic energy/momentum. The amplification of the force occurs because the area at position 2 is much larger so many more particles can impose a force there compared to the piston at position 1.
Imagine you impose a pressure $p$ at position $1$, by conservation of mechanical energy (disregard effects of gravity), this must be the same pressure at position $2$. So we can state:
$$p=\frac{F_1}{A_1}=\frac{F_2}{A_2}\rightarrow\frac{F_2}{F_1}=\frac{A_2}{A_1}$$
If we assume $A_2/A_1\gg1$ like in the figure it follows, $F_2/F_1\gg1$, i.e. a small force at position $1$ translates to a large force at position $2$.
By conservation of energy the amount of work $W$ at position $1$ must be the same as at position $2$, i.e.:
$$W=F_1\Delta x_1=F_2\Delta x_2\rightarrow\frac{\Delta x_2}{\Delta x_1}=\frac{F_1}{F_2}\ll1$$
We find that displacement of the piston at position $2$ is only a fraction of the displacement of the pistion at position $1$.
