Here I'll try to basically connect some dots to guide you through the example of the second text you posted...
Any quantum field theory of your choice associates certain integrals to observables, which you have to compute. The Feynman diagrams are representations of these integrals. The lines correspond to propagators, which encode the different field dynamics, and the vertices are expressions which contain the coupling strengths and the right amount of indices to connect your propagators. To derive the Feynman rules, you'll expand the integrals, read off the general structure and associate certain integrands to certain pictures. Then, with the rules in your pocket, you decide on a Feynman diagram of choice you want to compute, write down all the right terms and integrate over all the loose ends.
Now, you have an expression $ \mathcal{L}_{int} = g\,(\partial^{\mu}A )^2 B^2 $, which you identify as interaction term (there are two different fields after all) and you wonder what to do with the derivative, which you only know from the kinetic term. Well, to know what the propagators of the theories are, you need the whole Lagrangian/the full dynamics of the theory anyway, so this information will certainly get incorporated there. How the vertex expression turns out (your question) is what the second paper you posted is trying to describe:
If you derive what the Feynman rules are in momentum space, where the fields $A$ and $B$ get represented in terms of their Fourier modes ($A(x)=\int\text{d}p\, \hat A(p)\text{e}^{ipx}$), then you see that a derivative $\partial^\mu$ turns into a momnetum four-vector $p^\mu$ (under the intergral).
If you had the simpler interaction structure $g\, A ^2 B^2$, then your vertex would be typically represented merely by the number $g$ and the knowledge of which propagators end up there. Now, in deriving the Feynman rule for your specific problem which involves $g\, (\partial^{\mu}A )^2 B^2 $, your integrand will also contain a function of the momentum vector, e.g. $p^2$ from $$\partial^2 A(x)=\int \text{d}p\, \hat{A}(p)\text{e}^{ipx}\cdot p^2$$
Hence your vertex term (in momentum space representation), which is essentially the integrand without the propagator expressions (some denominators which look like "$\frac{1}{p^2+m^2}$" or so) will be not only "$g$" but something like "$g \cdot p^2$".
Clearly what this means is that the higher modes (big momenta etc.) might be dangerous objects, as you want your integrals to converge - you integrate over $p$, so higher powers in $p$ under the integral are usually not your friend. Very vaguely, if the direct coupling à la $S\sim g\int\text{d}x A^2B^2$ wants to be minimized then high $A$ means low $B$. From this perspective, a term "$S\sim g\int\text{d} x \, (\partial^{\mu}A )^2 B^2 $" makes you think "Oh, so the behaviour of the field $B$ doesn't only depend on the other field amplitude of $A$, but also directly on that fields relative local dynamics". but you really have to take a look at specific theories for specific implications.
If you look for physical (but more involved) examples, you can look up the Feynman diagrams of Yang-Mills theory (scrolling down a little on the page) and try to compare the interaction structure with all the vertices containing functions of momentum (the second and the last here).
Best Answer
The force is just the gradient of the potential energy. So it's not true that the energy is just force multiplied by $r$. It is the integral of the force w.r.t. $r$, which gives you the $1/r$ dependence.
Edit: here "potential energy" and "interaction energy" are used interchangeably.