That categorical definition is for pre-sheaves, the topological definition is for sheaves.
In topological pre-sheaves, a map is surjective if it is epimorphic for each open set $U$ in $X$.
In topological sheaves, however, we instead have to "sheaf-ify" the definition, and we say that the map is "surjective" if the sheaf-ification of the cokernel map is zero.
Basically, in both cases, you have two categories, $\mathcal{Sh}$ and $\mathcal{PSh}$, and in $\mathcal{PSh}$, the "surjective" maps are the ones that are epimorphisms on each $U$, but in the $\mathcal{Sh}$ catageory, you have a more complicated definition of "surjective" (or "epimorphism.")
Consider, instead, two categories, $\mathcal{Ab}$ the category of abelian groups, and $\mathcal{AbTF}$, the full subcategory of "torsion-free" abelian groups - that is, the abelian groups, $A$, where for any $n\in\mathbb Z$ and $a\in A$, $na=0$ iff $n=0$ or $a=0$.
There is the natural inclusion functor $\mathcal{AbTF}\to\mathcal{Ab}$ and a natural adjoint sending $A\to A/N(A)$ where $N(A)$ is the subgroup of nilpotent elements of $A$.
But in $\mathcal{AbTF}$, the "epimorphisms" are not the ones with cokernel (in $\mathcal{Ab}$) $0$, they are the ones with cokerkels which are nilpotent. So, for example, in $\mathcal{Ab}$, the morphism $\mathbb Z\to\mathbb Z$ sending $x\to 2x$ is not an epimorphism, that same map, when considered as a map in $\mathcal{AbTF}$, is an epimorphism.
So consider the "sheafification" functor $\mathcal{PSh}\to \mathcal{Sh}$ to be much like the functor $\mathcal{Ab}\to\mathcal{AbTF}$.
(I believe, but don't quote me, that $f:A\to B$ in $\mathcal{AbTF}$ is an epimorphism if and only if $f\otimes \mathbb Q:A\otimes \mathbb Q\to B\otimes\mathbb Q$ is an epimorphism in $\mathcal{Ab}$.)
The answer to 2) (and so 1) is yes. To see it, let's recall what the maps are.
Take a colimit cocone of $\mathcal{F}$ over $J_2$. Its restriction to $J_1$ is a cocone of $\mathcal{F}$ again. So it induces a unique morphism from $\mathrm{colim}_{J_1}(\mathcal{F}) \to \mathrm{colim}_{J_2}(\mathcal{F})$, this is a construction-free definition of the vertical map of your diagram.
The horizontal maps are constructed in the following way: given a natural transform $\alpha: \mathcal{F} \to \mathcal{G}$, take the colimit cocone of $\mathcal{G}$ and precompose it by the natural transform $\alpha$. By naturality of $\alpha$ this is a cocone under $\mathcal{F}$, and so it again induces a unique map from $\mathrm{colim}(\mathcal{F}) \to \mathrm{colim}(\mathcal{G})$.
Now compare the constructions of the two composites maps in your diagram. Let's look first at the one going down then left: i.e $\mathrm{colim}(\alpha) \circ \rho_F$ where $\rho_F$ is the restriction. To build the composite. First take the colimit cocone of $\mathcal{G}$ over $J_2$, precompose with $\alpha$, induce colimit, then restrict, then induce colimit again. By the universal property of the colimit, this map is in fact the unique map $\mathrm{colim}_{J_1}(\mathcal{F}) \to \mathrm{colim}_{J_2}(\mathcal{G})$ that makes the cocone obtained from the colimit cocone of $\mathcal{G}$ by precomposition by $\alpha$ and then restriction to $J_1$.
But precomposition by $\alpha$ and restriction to $J_1$ commute (in the sense that first restricting the diagram then precomposing with a restricted alpha is the same as precompose by $\alpha$ then restrict the whole thing.) So this map is also the map you get from the other composition, which you can check satisfies the same property because precompostion and restriction commute.
Best Answer
This is another proof of the isomorphism $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$, hence not really answer of the original question, but this was requested in the comments.
You have an adjunction $\operatorname{Hom}_{PSh(X)}(f^{-1}\mathcal{F},\mathcal{G})=\operatorname{Hom}_{PSh(Y)}(\mathcal{F},f_*\mathcal{G})$ and an isomorphism of functor (in fact a true equality here) $(g\circ f)_*=g_*\circ f_*$. Using this, we have : $$\begin{align*} \operatorname{Hom}_{PSh(X)}((g\circ f)^{-1}\mathcal{F},\mathcal{G})&=\operatorname{Hom}_{PSh(Z)}(\mathcal{F},(g\circ f)_*\mathcal{G})\\ &=\operatorname{Hom}_{PSh(Z)}(\mathcal{F},g_*f_*\mathcal{G})\\ &=\operatorname{Hom}_{PSh(Y)}(g^{-1}\mathcal{F},g_*f_*\mathcal{G})\\ &=\operatorname{Hom}_{PSh(X)}(f^{-1}g^{-1}\mathcal{F},\mathcal{G})\\ \end{align*} $$ This is natural in $\mathcal{F}$ and $\mathcal{G}$, hence by Yoneda, we have an isomorphism $(g\circ f)^{-1}\mathcal{F}\simeq f^{-1}g^{-1}\mathcal{F}$. Since this is natural in $\mathcal{F}$, we have an isomorphism of functors $(g\circ f)^{-1}\simeq f^{-1}g^{-1}$.