Since both limbs are open to the atmosphere, the atmospheric pressure can be ignored. In the initial configuration, the piston is in equilibrium when the hydrostatic pressures are balanced:
$$\rho_1 gh_1 S=\rho_2 gh_2 S;$$
where $h_1$ and $h_2$ are the heights of the liquid surfaces in the left and right limbs, respectively.
Let us consider that addition of the liquid of density $\rho_1$ into the left limb causes a shift of the piston at the distance $x$ to the right and the stiffness force appears. The equilibrium condition for the piston:
$$\rho_1 gH_1 S=\rho_2 gH_2 S+kx.$$
The shift of the piston at the distance $x$ cause an increasing by $x$ of the height of the upper surface of the liquid in the right limb. Due to the remaining of the difference in levels, the height of the upper surface of the liquid in the left limb also increases by $x$. Thus,
$$\rho_1 g (h_1+x) S=\rho_2 g(h_2+x) S+kx.$$
The equilibrium condition for the piston after addition of the liquid with density $\rho_2$ and volume $\Delta V$ to the left limb is:
$$\rho_1 g(h_1-x)S+\rho_2 gzS=\rho_2 g(h_2+x)S+kx.$$The upper surfaces are leveled when $z = \Delta h + 2x$. Therefore, using the value of $k$,
$$x=\frac{\rho_2 \Delta h}{2(\rho_1-\rho_2)}.$$
The volume of added liquid $\Delta V$ could be found as $\Delta V=S(\Delta h+2x)$.
Let us prove that if the piston is shifted by $x$ some volume of the liquid with density $\rho_1$ remains in the left limb. The equilibrium condition for the piston in the initial configuration:
$$\rho_1 gh_1 S=\rho_2 g(h_1+\Delta h) S \Rightarrow h_1=\frac{\rho_2 \Delta h}{\rho_1-\rho_2} \Rightarrow h_1>x.$$