Let $X$ be a topological space with a presheaf $\mathcal{F}$ and its sheafification $\varphi \colon \mathcal{F} \to \mathcal{G}$. The sheaf $\mathcal{G}$ is uniquely determined by the property that $\varphi_x \colon \mathcal{F}_x \to \mathcal{G}_x$ is bijective for all $x \in X$. Informally, $\mathcal{G}$ can be understood as the result of identifying all sections of $\mathcal{F}$ which agree locally, and of gluing all compatible sections of $\mathcal{F}$ together.
Although the above seems clear to me, I am not able to prove the following, which seems reasonable to me:
For any section $g \in \mathcal{G}(U)$ there is an open cover $U = \bigcup_i U_i$ and sections $f_i \in \mathcal{F}(U_i)$ such that ${f_i}_{|U_i \cap U_j} = {f_j}_{|U_i \cap U_j}$ for all $i,j$, and $\varphi_{U_i}(f_i) = g_{|U_i}$.
Is this statement even true?