Here is a simple answer: the collision rate can be said to be proportional to the ratio between velocity and the total surface area of the piston.

$PV=nRT$. Hence, $V\propto T$. The total volume is $2\pi r^2h$, so we can say $V\propto h$, where $h$ is the height of the column of gas.

Next, $KE=\frac{3}{2}kT$, where $k$ is Boltzmann’s constant and $T$ the temperature. Thus we can say $T\propto v^2$.

Hence, $h\propto v^2$, and from this relation, the collision rate $Z\propto \frac{h}{v}$. we notice that if $h$ decreases by a factor of $2$, $v$ would decrease by a factor of $\sqrt 2$, and the new ratio $Z\propto\sqrt 2\times \frac{h}{v}$. From which, we can see that collision frequency would increase.

Here is a simple answer: the collision rate can be said to be proportional to the ratio between velocity and the total surface area of the piston.

$PV=nRT$. Hence, $V\propto T$. The total volume is $\pi r^2h$, so we can say $V\propto h$, where $h$ is the height of the column of gas.

Next, $KE=\frac{3}{2}kT$, where $k$ is Boltzmann’s constant and $T$ the temperature. Thus we can say $T\propto v^2$.

Hence, $h\propto v^2$, and from this relation, the collision rate $Z\propto \frac{h}{v}$. we notice that if $h$ decreases by a factor of $2$, $v$ would decrease by a factor of $\sqrt 2$, and the new ratio $Z\propto\sqrt 2\times \frac{h}{v}$. From which, we can see that collision frequency would increase.

As to why there is a force, upon collision, there is change in momentum: impulse, which is the cross product of force and time.