So I found something here, and polished it a bit. Any feedback would still be appreciated :)
Let $\mathscr{F} \subset \mathcal{O}_X$ be a quasi-coherent
sheaf on $X$, and let $\mathscr{I}_1,\dotsc,\mathscr{I}_n$ be the ideal
sheaves associated to the irreducible and reduced compontents
$X_1,\dots,X_n \subset X$. Consider the filtration
$$
\mathcal{F} \supset \mathscr{I}_1 \cdot \mathscr{F} \supset \mathscr{I}_1
\mathscr{I}_2 \cdot \mathscr{F} \supset \dotsb \supset \mathscr{I}_1 \dotsb
\mathscr{I}_n \cdot \mathscr{F} = 0.
$$
The last equality iholds because $X$ is reduced, so $\mathcal{I}_1 \dots
\mathcal{I}_n = 0$.
Let $\mathscr{G}_k$ denote the $k$-th quotient of this filtration.
It is a quasi-coherent module on $X_k$, so its cohomology groups (on $X_k$)
vanish.
But if $\mathscr{G}$ is any quasi-coherent sheaf on $X_k \xrightarrow{i}
X$, then the cohomology groups of $i_* \mathscr{G}$ and $\mathscr{G}$ are
the same.
This is a special case of Exercise III 8.2, and can also be seen as
follows.
Take any injective resolution $0 \to \mathscr{G} \to \mathscr{I}^*$.
Because $X_k$ is a subspace of $X$, the push-forward is exact in this
situation:
The stalks at points in $X_k$ are the same, and the stalks outside of $X_k$
are all $0$.
So $0 \to i_* \mathscr{G} \to i_* \mathscr{I}^*$ is a resolution, and the
$i_* \mathscr{I}^*$ are flasque, so we can use them to compute cohomology.
But applying $\Gamma(X, \cdot)$ just reproduces $\Gamma(X_k,
\mathscr{I}^i)$, so $H^i(X, i_* \mathscr{G}) = H^i(X_k, \mathscr{G})$.
If we apply the long exact sequence to the sequences
$$
0 \to \mathscr{I}_1 \dotsm \mathscr{I}_{k-1} \cdot \mathscr{F} \to
\mathscr{I}_1 \dotsm \mathscr{I}_{k} \cdot \mathscr{F} \to
\mathscr{G}_k \to 0,
$$
we get a surjection $H^1(X, \mathscr{I}_1 \dotsm \mathscr{I}_{k-1} \cdot
\mathscr{F}) \to H^1(X, \mathscr{I}_1 \dotsm \mathscr{I}_{k} \cdot
\mathscr{F}) \to 0$. Composing the surjections for all $k$ we get a surjection $0 = H^1(X, \mathscr{I}_1\dotsm\mathscr{I}_n \cdot \mathscr{F}) \to H^1(X, \mathscr{F})$, hence $H^1(X, \mathscr{F}) = 0$.
Pick a finite number of generators of $M_i$ as an $A_i$ module. Since the restrictions of global sections generate $M_i$, we can write each generator of $M_i$ as a finite $A_i$-linear sum of the global generators (this works because we can only consider finite sums!). Thus we get a finite list of global generators which suffice to generate $M_i$. Now take the union of the lists for each $M_i$ across all elements of your finite cover, and you get a finite list of global generators for your global sheaf.
Best Answer
2024-02-20: There is an important error in your proof (which I should have pointed out earlier): the claim that "sheafifying injective $A$-modules yields injective $\mathcal{O}_X$-modules" isn't true. Take for example $A=\Bbb F_2$ and $M=\Bbb F_2$ - then $M$ is injective, but $\widetilde{M}$ is not injective as a sheaf of abelian groups on $\operatorname{Spec} A=\{pt\}$ because it is not divisible. $\widetilde{M}$ is an injective object in the category of quasi-coherent sheaves, though, but that doesn't help because the various Ext functors here are defined out of $\mathcal{O}_X$-mod.
I suspect the answer is that one actually does need finite generation here for $\operatorname{Ext}^i_X(\widetilde{M},\widetilde{N})\cong \operatorname{Ext}^i_A(M,N)$, but I do not have a counterexample at hand.
2021-12-07: $M$ finitely generated is necessary for $\mathcal{E}xt$, which is the second half of the problem.
To see that $M$ finitely generated is necessary for the second half of the problem, first observe that Hom only commutes with localization when the first argument is finitely presented. For instance, taking $A=N=\Bbb Z$ and $M=\Bbb Z[\frac12]$, $\operatorname{Hom}_A(M,N)=0$ but the Hom sheaf is nonzero over $D(2)\subset\operatorname{Spec} \Bbb Z$, so we get a counterexample involving $\operatorname{Ext}^0$.