Sheafification of a subpresheaf

I am assuming $\mathscr H$ is a sheaf of abelian groups on a topological space $X$.
Since $\mathscr H$ is a sheaf, by the universal property of sheafi-fication, the morphism of pre-sheaves $\iota:\mathscr F\hookrightarrow\tilde {\mathscr F} $ factors through $\mathscr F\xrightarrow{j}\mathscr F^{sh}\xrightarrow{\bar \iota }\tilde{\mathscr F}$. To see that $\bar \iota: \mathscr F^{sh}\rightarrow \tilde {\mathscr F}$ is an isomorphism, it is sufficient to check at the level of stalks.
Observe that at the level of stalks, we have for $x\in X$, the base space, the composition $$\mathscr F_x\xrightarrow{j_x}\mathscr F^{sh}_x\xrightarrow{\bar \iota_x }{\tilde{\mathscr F}}_x=\mathscr F_x $$ which is identity. Moreover, $j_x$ is an isomorphism. This comes from the very construction of the sheafification. Thus $\bar \iota_x$ is an isomorphism for every $x\in X$ and hence $\bar \iota $ is an isomorphism.

Note: The standard construction of $\mathscr F^{sh}$ is to consider $\mathscr F$ as a sub-presheaf of the sheaf of stalks $\mathscr H(U):=\{\prod_{x\in U}s_x: s_x\in \mathscr F_x \}$ and then look at $\tilde {\mathscr F}$ in this sheaf.

Given a sheaf $\mathscr{G}$ and a morphism of presheaves $f:\mathscr{F}\to \mathscr{G}$, you can directly construct a morphism of presheaves (and therefore of sheavs) $\widetilde{f}:\widetilde{\mathscr{F}}\to \mathscr{G}$ that extends $f$. Given a section $s\in \widetilde{\mathscr{F}}(U)$, choose any open cover $\{U_i\}$ of $U$ such that the restrictions $s\mid_{U_i}$ lie in $\mathscr{F}$. Then you can apply $f$ to each of these restrictions, and glue their images to get $\widetilde{f}(s)\in \mathscr{G}(U)$. It is not hard to check that this is well-defined, i.e. independent of the open cover, and that in fact $\widetilde{f}$ is the unique extension of $f$. The latter shows that $\widetilde{\mathscr{F}}$ is a sheafification.

Regarding the functoriality you mention, since your definition requires choice of a sheaf $\mathscr{H}$, I am not sure how to make sense of this (of course you know after the fact that sheafification is functorial by the universal property).