There's a functor that taakes a presheaf $\mathcal F$ on $X$ and assigns to it the stalk at $x$, written $\mathcal F_x$. There's also a result saying that this functor is exact. In proving this, we have to deal with the kernel and the image of maps $\mathcal F_x\to\mathcal G_x$ in the category of sets. If I don't want to deal with the genral notion of kernels/images in abelian categories, is it right to think of kernels and images in the category of sets in the most natural way, namely as of the set of elements sent to zero and the set of elements that come from something, respectively?
And a related question (if what I said above makes sense), what is "zero" in $\mathcal G$? If we're dealing with with the presheaf of some functions, I suppose it is the equivalence class of the pair $(U,0)$, where $0$ is the zero function. But what is it for general presheaves?
The relevant categories in your case are not abelian, so your interpretation of the meaning of exact functor is incorrect. A functor is exact if it preserves finite limits and finite colimits. The functor you are looking at has a simple description: $\mathcal F_x = \mathrm{colim} (F{\mid}_x)$, namely the colimit of the restriction of $\mathcal F$ to neighbourhoods of $x$.You should now be able to check the properties of this functor.