Question: Potential energy can be thought as the amount of work that the force can potentially do on the point because of its position. $$W=-\Delta U=U\_{initial}-U\_{final}$$

A positive work done by a force translates into a negative variation of potential energy. That sounds ok, given the interpretation of $U$ stated above. If a force does some work, then the "potentiality" of doing more must decrease.

But the equation says also that any time the force does a negative work, the potential energy increases. Why does this happen, in the light of such interpretation of $U$?

Answer: 
If a force does **negative work**, it is in fact trying to work against another force, doing **positive work**.

When you lift up a book from the floor, gravity does *negative work* on the book, while you do *positive work*. And the books rises higher up, so $U$ increases.

* Negative work just means "**receiving**" instead of "**giving**" away energy. Which basicly is the same as saying that *something **else** is doing work on it*.

There is not more to it than that.
It's a convention. The real reason is so that we can have:

$$\begin{align}
\Delta {\rm KE} &= W\_{\rm ext}\\
\Delta {\rm KE} &= W\_{\rm con} + W\_{\rm noncon}\\
\Delta{\rm KE} &= -\Delta {\rm PE} + W\_{\rm noncon}\\
\Delta{\rm KE} + \Delta {\rm PE} &= W\_{\rm noncon}\\
\Delta {\rm E} &= W\_{\rm noncon}
\end{align}$$

Which doesn't work without the sign choice. So it's really just an arbitrary convention to make it so you can add potential energy to kinetic energy in the work-energy theorem.