I was wondering if the following elementary statement about sheaves is true: If $\mathcal{F}$ is a subsheaf of $\mathcal{G}$ (both are sheaves on $X$, a topological space) with stalks we denote by $\mathcal{F}_P$ for each $P \in X$. Then is $F$ isomorphic to the sheaf $\mathcal{F}^{\dagger}$ whose sections $\mathcal{F}^{\dagger}(U)$ are defined to be the group of functions $f: U \rightarrow \amalg \mathcal{F}_P$ such that for all $P \in X$, $f(P) \in \mathcal{F}_P$ and moreover, there is some open neighborhood of $P$, call it $V \subset U$ and section $s \in \mathcal{G}(V)$ such that $f(Q) = s|{Q}$ for all $Q \in V$.
So informally, given a subsheaf $\mathcal{F} \subset \mathcal{G}$, is $\mathcal{F}$ isomorphic to the sheaf formed by taking collections of germs of $\mathcal{F}$ that are compatible with respect to $\mathcal{G}$?
Yes. In particular, if we set $Q := P$, then for each $P \in U$ we get an open neighborhood of $P$, $V \subseteq U$, and section $s \in \mathcal{G}(V)$ such that $f(P) = s|P$. By definition of stalks, this implies that for some other open neighborhood of $P$, $V' \subseteq V$, we have $s|_{V'} = t|_{V'}$ where $t$ is a representative of $f(P) \in \mathcal{F}_P$. Thus, we see that $f$ satisfies the compatibility condition required for $f$ to induce an element of $\mathcal{F}(U)$.
A closely related fact is: for any $g \in \mathcal{G}(U)$, we have that $g \in \mathcal{F}(U)$ if and only if $g_x \in \mathcal{F}_x$ for each $x \in U$. (And in general, $\{ x \in U \mid g_x \in \mathcal{F}_x \}$ is an open subset of $U$. This relates to the notion of the subobject classifier $\Omega$, which is a sheaf of sets such that $\Omega(U) = \{ V \subseteq U \mid V \mathrm{~open} \}$, and the relation between subsheaves of $\mathcal{G}$ and morphisms of sheaves $\mathcal{G} \to \Omega$.)