Let $X$ be a noetherian scheme. I'm going through the proof that to check that a quasi-coherent $\mathcal{F}$ is coherent, it suffices to check that $\Gamma(U_i, \mathcal{F})$ is a finite $\Gamma(U_i, O_X)$-module for the open sets $U_i$ of one affine covering $\{ U_i \}$ of $X$.
By some results on quasi-coherent modules, we get that if $U$ is any affine, $M = \Gamma(U, \mathcal{F})$, and $M_i = \Gamma( U \cap U_i, \mathcal{F})$, then $M_i$ is a finite $\Gamma(U \cap U_i, O_X)$-module. Cover $U$ by a finite set of $U \cap U_i$'s, say $U \cap U_1, \ldots, U \cap U_n$. Then it says "to generate $M$ over $\Gamma(U, O_X)$, it suffices to take enough elements of $M$ so that their images in each $M_i$ generate $M_i$ over $\Gamma(U \cap U_i, O_X)$."
I don't follow this last sentence, because this seems to only make sense if we know that $\Gamma(U, \mathcal{F}) \to \Gamma(U \cap U_i, \mathcal{F})$ is surjective, which doesn't always seem to be the case... Any clarification is appreciated. Thank you.
Set $R=\Gamma(U,\mathcal{O}_X)$ and $R_i=\Gamma(U \cap U_i, \mathcal{O}_{X})$. Recall that if $\mathcal{F}$ is quasi-coherent then $\mathcal{F}|_U$ is the sheaf associated to your $M$ (you are assuming $U$ is affine).
Since any Noetherian scheme is quasi-separated, you have $U \cap U_i= \bigcup_j U_{f_{i,j}}$ for a finte number of $f_{i,j} \in \Gamma(U, \mathcal{O}_X)$. Note that $M_{f_{i,j}}= \Gamma(U_{f_{i,j}}, \mathcal{F})$ is finitely generated over $R_{f_{i,j}}$ beacause of your hypothesis and the fact that $U_{f_{ij}}$ is an affine open subset of $U_i$. Pick a finte number of generators $m_{i,j,k}/f_{i,j}^{n_k} \in M_{f_{i,j}}$, then $\{m_{i,j,k}\}_{k} \subset M$ are such that their image generates $M_{f_{i,j}}$ over $R_{f_{i,j}}$.
At this point you could conclude directly that $\{m_{i,j,k}\}_{i,j,k}$ generate $M$ beacuse $U= \bigcup_{i,j} U_{f_{i,j}}$.
I think you should also be able to deduce that $\{m_{i,j,k}\}_{j,k}$ generates $M_i$ if you like.