You know that if you have a point charge with charge $Q$, then the potential difference $V$ between spatial infinity and any point a distance $r$ from the charge is given by $$V_\textrm{point}=\frac{kQ}{r}.$$ You also know that the electric field from an infinite sheet of charge with charge density $\sigma$ is given by $$E_\textrm{sheet}=2 \pi k \sigma. $$
Because the electric field is uniform, you correctly concluded that there must be an infinite potential difference between any point and spatial infinity. You are surprised because this seems at odds with the first formula for $V_\textrm{point}$.
However, there is a good explanation. If $\dfrac{kQ}{r}$ is originally for a point charge, what values of $Q$ and $r$ should we plug in for the case of a sheet? Well, notice that the sheet has an infinite amount of charge, so that perhaps $Q$ should be infinite. This explains why we might get an infinite potential difference. However, there is a competing effect occuring with $r$. As you go farther out on the infinite sheet, you get farther and farther away from the point where you are trying to compute the potential, so it seems like maybe $r$ should be very big, maybe infinitely big as well. Let's see how to do the problem correctly.
To do the problem correctly, you need to realize that each point on the infinite sheet acts like a little point charge, so each point gives its own $\dfrac{kQ}{r}$ contribution. The total potential, by superposition, is the sum of these contributions. We can sum up the contributions by integration. Let's first pick a coordinate system where the plate is on the $x$-$y$ plane, and the point where we want to know the potential is on the $z$ axis. We can switch to cylindrical coordinates where $\rho = \sqrt{x^2+y^2}$. Then the distance $r$ between the point with coordinate $z$ on the $z$ axis and a point with coordinate $\rho$ is given by $r = \sqrt{z^2 + \rho^2}$, and so, applying the $kQ/r$ formula, the contribution $dV$ to the potential from a bit of charge $dQ$ a distance $\rho$ from the origin is given by $$dV = \frac{kdQ}{\sqrt{z^2+\rho^2}}.$$ Integrating this over all $\rho$ we find
$\begin{equation}
\begin{aligned}
V&=\int^\infty_0 \frac{2 \pi k \sigma \rho d \rho}{\sqrt{z^2+\rho^2}} \\
&= \pi k \sigma \int^\infty_0 \frac{du}{\sqrt{z^2+u}}\\
&=2 \pi k \sigma \left( \sqrt{\infty + z^2} - |z| \right).
\end{aligned}
\end{equation}$
Because of the infinity in the square root, the potential above is in fact infinite, even though were started with a finite $kQ/r$ law. This infinity was possible because we had infinitely much $Q$. Notice the electric field still works out because the infinite part does not have a spatial gradient: $$E=-\dfrac{dV}{dz} = -2 \pi k \sigma \left( \dfrac{z}{\infty + z^2} - 1\right) \hat{z} = 2 \pi k \sigma \hat{z}.$$
The difference between electric potential at two points is proportional to the work done by moving charged particle in a present field. The work is the force multiplied by the distance: $F \cdot s$, or more generally $\Delta U \propto \int_P \vec{F} \cdot d\vec{s}$.
Given that the force scales as $\frac{1}{r^2}$ it should be easy to see why multiplying it by distance in units of $r$ we expect to get a potential that scales as $\frac{1}{r}$.
This process works in reverse too, the field (which is proportional to the force on a particle) is basically a derivative of the potential: $\vec{E} = -\vec{\nabla} U$. So changing potential in a region will result in force acting on a charged particles in this region.
For energy transfer - take a look at Poynting vector. It describes the flux of energy through the area.
Best Answer
For a constant force, we can obtain the work by multiplying the force by the distance it is applied over.
If we however have a non-constant force, then the work is $W = \int \vec{F} d\vec{s}$.
In your case, this translates to $W = - \int_{\infty}^{r} \frac{KQ_2}{r'^2} dr'={\frac{KQ_2}{r'}}|^r_{\infty}=\frac{KQ_2}{r}$.
So the fact that you obtain the right answer needs to be called chance because you are not calculating your work correctly.