Consider $X$ a scheme, $\mathcal F, \mathcal G$ globally generated sheaves. I want to show that $\mathcal F\otimes \mathcal G$ is globally generated.
By definition we have surjective morphisms $f:\oplus_{I\in I}\mathcal O_X\to \mathcal F$, $g:\oplus_{j\in J}\mathcal O_X\to \mathcal G$. From them I considered for $U\subseteq X$ open $(\oplus_{I\in I}\mathcal O_X)\otimes(\oplus_{j\in J})(U)\to \mathcal F(U)\otimes \mathcal G(U)$ given by $u\otimes v\mapsto f(u)\otimes g(u)$
By considering stalks I think this is surjective (we can't deduce surjectivity from maps on open sets). So we have a surjective morphism $\varphi$ from $(\oplus_{I\in I}\mathcal O_X)\otimes(\oplus_{j\in J}\mathcal O_X)$ to the tensor presheaf. However we want a map which goes in the tensor sheaf, which have the same stalks. At first I considered the sheafification map $\theta: ( F\otimes \mathcal G)^{\operatorname{pre}}\to \mathcal F \otimes \mathcal G$ and said $\theta_p\circ \varphi_p=\varphi_p$ but reading it again made me realize it could be false, so $\theta \circ \varphi$ might not being a good choice despite being the most natural way to have a map to the sheaf.
Is there a way to make this work ?
If $X$ is a ringed space and $\mathcal{F}$ is a sheaf of $\mathcal{O}_X$-modules, then the functor $-\otimes_{\mathcal{O}_X} \mathcal{F}$ is a right exact functor on the category of $\mathcal{O}_X$-modules. This immediately resolves your issue: if $\mathcal{O}_X^{\oplus I} \to \mathcal{F}\to 0$ and $\mathcal{O}_X^{\oplus J} \to \mathcal{G}\to 0$ are surjections exhibiting $\mathcal{F}$ and $\mathcal{G}$ as globally generated, applying $-\otimes_{\mathcal{O}_X} \mathcal{F}$ to $\mathcal{O}_X^{\oplus J} \to \mathcal{G}\to 0$ gives a surjection $\mathcal{F}^{\oplus J} \to \mathcal{F}\otimes_{\mathcal{O}_X}\mathcal{G}\to 0$, which when concatenated with the surjection $\mathcal{O}_X^{\oplus(I\times J)} \to \mathcal{F}^{\oplus J}$ coming from taking the direct sum indexed by $J$ of the surjection $\mathcal{O}_X^{\oplus I}\to\mathcal{F}$ yields a surjection $\mathcal{O}_X^{\oplus (I\times J)}\to\mathcal{F}\otimes\mathcal{G}$.
To se the claim that tensor is right-exact on an arbitrary ringed space, one shows that the stalks of the tensor product presheaf (and hence the tensor product sheaf) are $\mathcal{F}_x\otimes_{\mathcal{O}_{X,x}} \mathcal{G}_x$, and then one uses that the tensor product of modules is right exact. See here for details, for instance.