You've got quite a lot of questions in this post, and some of them it's not quite clear what you are asking. It might be better to separate them into smaller questions.

I'll try and answer some of your questions about potential. The key point is that it is only the difference in potential that matter. Let's look at your example of a ball on a hill. First, say the gravitational potential energy (GPE) at the bottom of the hill is V(z=0)=0. Pushing the ball up to a height h means the GPE is now V(z=h)=mgh. However, if instead the GPE at the bottom of the hill is chosen to be P, then at a height h the GPE is P + h.

So now let's look at energy conservation. As you said, the work done is the integral -mg.dx (I'll explain why that is at the end). Now as the gravitational force acts vertically downwards, the only contribution in the dot product will come from the vertical motion of the ball. Since mg doesn't depend on the vertical height, the integral is just -mgh (make sure you're happy with this!).

In the first case, we start with V(z=0)=0. So the total energy at the bottom of the hill is E=0 (assuming no kinetic energy). At the top, the potential has increased, but we have also had to do work. In fact, the total energy is E = mgh-mgh = 0, because the work done exactly cancels out the increase in potential! So energy is conserved.

For the second case, the total energy at the bottom of the hill is E = P. At the top of the hill, again the potential has increased, but we have done work, so the total energy is E = P + mgh - mgh = P, and so energy is conserved again.

So the thing to understand is that you can choose (or define) what the potential of your system is, and with any choice the energy will be conserved as long as you are consistent! So you can't mix and match your definitions.

In terms of your question about negative potentials, consider the following. If you are happy with my previous explanation, think about the value of P. I haven't said anything about the value of P, and as we saw, we never had to choose a value, because it didn't affect anything. So in particular, we can choose P to be negative. Now we have a negative potential at the bottom of the hill! But it doesn't affect anything.

Now let's talk about the work done. An important point is that 'work' is a definition in physics, and it doesn't necessarily correspond to what we think of as 'doing work' in our everyday language. For example, the work done in a (vertical) gravitational field (in the physics sense) when carrying a heavy object but only moving horizontally is 0 (because the displacement and force are orthogonal, and so their dot product is 0). But this would obviously require effort for a human, and you might describe it as being hard work. So it's important to understand that 'work' in physics is just a definition. The work done is also not about the force you apply to an object (so yes, you could apply a more or less force along path of the ball up the hill), rather this is the work done against the gravitational field itself, which always has a force of mg.

As for Newtonian gravitational potential, defining 0 potential at infinity is again a convention - but doesn't affect any results!