Analysis of the effect of force $F$ on the system
I assume that $A_i$ is the surface area of piston $i$, for $i=1,2$.
If the force $F$ pushes piston 1 down by some height $h_1$, there is a pression increase
on $p_1$ due to the force $F$ applied to surface $A_1$, and a decrease
due to the loss of the water level above $p_1$ of height $h_1$, which is $\rho g h_1$.
Thus the pressure in $p_1$ is increased by $F/A_1-\rho g h_1$.
Whatever happens, the pressure will be
the same in $p_1$ and $p_2$ since they are at the same height. The
same is true for $p_3$ and $p_4$.
(Note that what matters is height, not depth, if you can have
different water levels, or different pressures at the top)
Thus the same pressure increase occurs in $p_2$. But the pressure in
$p_2$ is balanced by an increase $h_2$ in water level, i.e. a raising of
piston 2, since there is no force on piston 2. This pressure increase
is $\rho g h_2$.
Hence we have $F/A_1-\rho g h_1=\rho g h_2$ ,
i.e., $F/(\rho g)=A_1 h_1 + A_1 h_2$
(Remember that $h_1$ and $h_2$ are measured in opposite directions.)
Now, since the amount of water stays the same, the volume lost on one
side equal the volume gained on the other side: $h_1 A_1=h_2 A_2$.
Hence $F/(\rho g)=h_2 A_2 + A_1 h_2$ ,
i.e., $h_2=F/(\rho g (A_1+A_2))$.
and $h_1=h_2 A_2/A_1= F A_2/(A_1 \rho g (A_1+A_2))$
The pressure change
The pressure under piston 2 was initially the atmospheric pressure.
As the water level rises by height $h_2$, the pressure at this initial
level is increased accordingly by $h_2 \rho g$, which is equal to
$F/(A_1+A_2)$ (see computation of $h_2$ above).
Then, according to Pascal's law, the increase in pressure is
the same everywhere in the liquid, so that everywhere the pressure
is increased by $F/(A_1+A_2)$.
This is true in particular for the 4 points under consideration. But the
pressure differential between any two points stays of course the same.
A minor additional remark
This reasonning, including the use of Pascal's law, assume that water is not compressible, which is not
absolutely true, though it can generally be neglected, except for extremely high pressures.
The pressure increase of $F/(A_1+A_2)$ does cause a very minute compression of the fluid,
resulting in a minute increase of specific mass (density) $\rho$. Hence the
pressure difference between $p_1$ and $p_3$ is actually very slightly
As I said, this is negligible for water in most situations, but it could be more
significant for a more compressible fluid. If more significant, it
should also be accounted for in writing the equations above,
particularly regarding volume of fluid which would be reduced.