That's pretty much it. Your example, you didn't specify the units for charges. If the proton (charge +e) were (distance x) mid-way between a charge of +6e and -10e then the net potential is given by:
[tex]V=\frac{6ke}{x}-\frac{10ke}{x} = -\frac{4ke}{x}[/tex]
You should be moving away from a force approach to motion and more towards using energy instead. There will be a transition where you will find forces easier to visualize.
The proton is acted on by a net force, the total work is associated with this net force via W=F.d (note that force and displacement are vectors?)
You can expand the F term into each individual force, and each force into components along and perpendicular to the displacement, getting a big sum ... each term of the sum will, itself, be work (energy) and you'll see how some of the terms will be positive and some negative. Thinking in terms of energy becomes very convenient later.
If you imagine a system of two charges, and you move a proton to a position where one charge contributes +6V and the other contributes -10V then the proton, by virtue of being in that position, has -4eV potential energy[1].
If you then move it into a position where is has -7eV potential energy then it has gained 3eV kinetic energy. If you wanted to return it to where it has 0eV potential energy, you'd have to give the proton 7eV of kinetic energy (somehow) to get it there.
You can imagine the same thing with gravity ... gravitational potential energy [close to the Earth is approximated by] mgh, where h is measured upwards from the ground ... this can be a negative number, for instance, if your mass is down a hole.
For this reason, regions of negative potential or often called "potential wells". (Regions of positive potential are called "barriers".)
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[1] spot the units ... eV = electron-volt: convenient for protons which have the charge of a single electron only positive. I could have expressed potential as J/e instead of Volts which is J/C and then expressed energies as Joules.