Let $\mathscr F$ be a sheaf on $X$. For any $p\in V$ ($V$ is an open subset of $X$), there is a map $$(-)_p: \mathscr F(V)\to \mathscr F_p\\ s\mapsto s_p=[(V, s\in \mathscr F(V))]$$
Now suppose $V$ is covered by open $\{V_j\}$. Let $p\in V_i\cap V_j$ and let $f_i\in \mathscr F(V_i)$. Does the statement $$(f_i)_p=(f_j)_p$$ imply $$(Res_{V_i, V_i\cap V_j} (f_i))_p=(Res_{V_j, V_i\cap V_j} (f_j))_p?$$
The first statement literally means $[(V_i, f_i\in \mathscr F(V_i))]=[(V_j, f_j\in \mathscr F(V_j))]$, which is true iff there is some $W\subset V_i\cap V_j$ containing $p$ s.t. $Res_{V_i,W} (f_i)=Res_{V_j,W}(f_j)$. Taking stalks at $p$ would give almost what is required, but not exactly (we can't say that the previous equality holds for $W=V_i\cap V_j$, right?). Then is the implication I'm trying to prove not valid? Or if it is, how to prove it?
(If this matters, I came across this problem while trying to prove that a tuple of compatible germs $(s_p)_{p\in V}$ always lies in the image of $\mathscr F(V)\to \prod_{p\in V} \mathscr F_p$.)
Note that \begin{align*} \mathrm{Res}_{V_i, W}(f_i) &= \mathrm{Res}_{V_i \cap V_j, W}(\mathrm{Res}_{V_i, V_i \cap V_j}(f_i)) \\ \mathrm{Res}_{V_j, W}(f_j) &= \mathrm{Res}_{V_i \cap V_j, W}(\mathrm{Res}_{V_j, V_i \cap V_j}(f_j)) \end{align*} so if $W$ is as you mentioned in the question, we can write $$\mathrm{Res}_{V_i \cap V_j, W}(\mathrm{Res}_{V_i, V_i \cap V_j}(f_i)) = \mathrm{Res}_{V_i \cap V_j, W}(\mathrm{Res}_{V_j, V_i \cap V_j}(f_j))$$ which implies that $(\mathrm{Res}_{V_i, V_i \cap V_j}(f_i))_p = (\mathrm{Res}_{V_j, V_i \cap V_j}(f_j))_p$.