Let $X$ be a topological space. We work with the category which has its open sets as objects and inclusions as morphisms and for a presheaf $F$ on this and $x \in FU$ we denote $x \upharpoonright V$ for $F(V \subseteq U)(x)$. We say a sieve $S$ on $U$ is covering if $\bigcup S = U$.
For a presheaf $F$ and the associated local homeomorphism $\pi: \coprod_{x \in X} G_x \to X$ the following holds: every morphism of presheafs $\mu: F \to H$ where $H$ is a sheaf factors uniquely through the sheaf corresponding to $\pi$ (under the equivalence between the full subcategory of $Top/X$ of étale maps and sheaves).
Let us call the associated sheaf $\tilde{X}$ which is defined to be $\tilde{X}(U) = \{s: U \to X \ | \ \pi s = id, \ s \textrm{ is continuous}\}$. I defined $p: F \to \tilde{X}$ as follows:
$p_U(\alpha) = (x \mapsto (x, \alpha_x))$ where with $\alpha_x$ I mean the germ of $(\alpha, U)$ at $x \in U$. After a bit of checking we see that $p_U$ is a local section of $\pi$ and that they collect into a natural transformation.
We want to factor $\mu$ as $\tilde{\mu} \circ p$ for some presheaf morphism $\tilde{\mu}$. That means we have to define $\tilde{\mu}_U(p_U(\alpha)) = \mu(\alpha)$.
I hope this uniquely determines $\tilde{\mu}_U$ on all elements of $\tilde{X}(U)$ but have been unable to proof this.
The approach I think is most obvious is to find a way to write any local section $s$ as the amalgamation of some compatible $(x_f \ | \ f \in S)$ family consisting out of elements of the form $p_U(\alpha)$. Then $(\tilde{\mu}(x_f) \ | \ f \in S)$ is also a compatible family. Must then $\tilde{\mu}_U(s)$ be equal to the amalgamation of that compatible family in order for $\tilde{\mu}$ to be a natural transformation?