When we apply a 'force' onto a surface, is there a 'pressure distribution'?

The piston will flex or temporarily deform slightly so as to distribute the force applied by your finger across its whole horizontal area. If the piston is not strong enough or rigid enough to do this then it will permanently deform (imagine hammering a nail into wood) or break (imagine hammering a nail into a sheet of glass).

First of all, you consider the piston (mass $m_{p}$) as a rigid body which is constrained to move inside the cylinder. Let's call $S$ the surface of the piston. Now you put above the piston, on its center, a cylindrical weight with mass $m$ and surface $S_{w}$. At the equilibrium (when the piston is firm), the force equation on the vertical axe, gives you: $$(m_{p}+m)g + P_{0} S=P_{gas}S$$ Where $P_{gas}$ is the pressure of the gas inside the cylinder at the equilibrium (you can measure it directly) and $P_{0}$ is the atmospheric pressure. $P_{gas}$ acts on every point of the low side of the piston. While on the top side, acts the atmospheric pressure and under the weight, but only there, the pressure is: $$P_{w}=frac{mg}{S_{w}}+ P_{0}$$ For example, to understand the magnitudes, if you consider $m=10kg$, you have: $$P_{w}^{(bar)}≈frac{10^{3}}{S_{w}^{(mm^{2})}}+ 1$$ Which is the function rapresented below:

So if $S_{w}=10mm^{2}$ the pressure on those points is about $100bar$ (extremely high).