Ah, I just realized: $\mathcal F \overset{\mathrm{def}}{=} \mathcal G$.
You basically want to construct a morphism $\Gamma(V, \varphi^{-1} \mathcal F) = \varinjlim_\limits{U \supseteq \varphi(V)} \Gamma(U,\mathcal F) \to \varinjlim_\limits{U' \supseteq \varphi(V')}\Gamma(U', \mathcal F) = \Gamma(V', \varphi^{-1}\mathcal F)$ for any inclusion $V' \subseteq V$ of open subsets.
Now the universal property of the colimit used to compute $\Gamma(V,\varphi^{-1}\mathcal F)$ tells you how to obtain morphism starting at $\Gamma(V,\varphi^{-1}\mathcal F)$.
Namely: You have to give a morphism $\Gamma(U,\mathcal F) \to \varinjlim_\limits{U' \supseteq \varphi(V')}\Gamma(U', \mathcal F) = \Gamma(V', \varphi^{-1}\mathcal F)$ for any open $U \supseteq \varphi(V)$ compatible with the transition maps (which are of course given via restriction).
Now just note that since $V' \subseteq V$ one certainly has $U \supseteq \varphi(V) \supseteq \varphi(V')$ so that $\Gamma(U,\mathcal F)$ is a part of the second colimit, i.e. comes with a canonical map into $\Gamma(U,\mathcal F) \to \Gamma(V',\varphi^{-1}\mathcal F)$ as desired.
Then you have to check that these maps are compatible with restrictions so that they induce the desired morphism $\Gamma(V,\varphi^{-1}\mathcal F) \to \Gamma(V', \varphi^{-1} \mathcal F)$.
$\textit{Uniqueness}$ of the morphism induced on the colimit then tells you that this construction behaves well with respect to inclusions $V'' \subseteq V' \subseteq V$, i.e. is functorial.
Best Answer
I gave it a go, and I think I might have solved it, but I only know the basics on sheaves so you should definetely double check my proof.
we have that $ \mathcal{F}_{x} = \mathcal{F}(\{x\}), \ \mathcal{G}_{x} = \mathcal{G}(\{x\}), \ \mathcal{H}_{x} = \mathcal{H}(\{x\}). \ $ Since $\ \ \large0 \rightarrow \mathcal{F} \rightarrow \mathcal{G} \rightarrow \mathcal{H} \rightarrow \large0 \ $ is exact, also $ \ \large0 \rightarrow \mathcal{F}_{x} \rightarrow \mathcal{G}_{x} \rightarrow \mathcal{H}_{x} \rightarrow \large0 \ $ is exact.
In order to prove that $ \ \ \large0 \rightarrow i_{\star}(\mathcal{F}) \rightarrow i_{\star}(\mathcal{G}) \rightarrow i_{\star}(\mathcal{H}) \rightarrow \large0 \ $ is exact, we can just prove that it is exact on the stalks. But this is indeed the case:
Let $y \in X \ \ \text{s.t. there's an open set} \ U \subset X \ \text{with } x \notin U.$ Then $i_{\star}(\mathcal{F})_{y} = i_{\star}(\mathcal{G})_{y} = i_{\star}(\mathcal{H})_{y} = \large0$
If $ \ \ \forall \ \ U \subset X \text{ such that } y \in U \text{ we also have } x \in U, \ $ then $ \ i_{\star}(\mathcal{F})_{y} = \mathcal{F(\{x\})}, \ i_{\star}(\mathcal{G})_{y} = \mathcal{G(\{x\})}, \ i_{\star}(\mathcal{H})_{y} = \mathcal{H(\{x\})}$.
In both cases the sequence $ \ \ \large0 \rightarrow i_{\star}(\mathcal{F})_{y} \rightarrow i_{\star}(\mathcal{G})_{y} \rightarrow i_{\star}(\mathcal{H})_{y} \rightarrow \large0 \ $ is exact. Therefore $ \ \ \large0 \rightarrow i_{\star}(\mathcal{F}) \rightarrow i_{\star}(\mathcal{G}) \rightarrow i_{\star}(\mathcal{H}) \rightarrow \large0 $ is also exact.