Why tensor product of two sheaves of modules is a sheaf of modules?

If $(X,\mathcal O_X)$ is a scheme and if $U\subset X$ is is an open affine subscheme, then we have for all quasi-coherent sheaves $\mathcal F,\mathcal G$ of $\mathcal O_X$-modules the extremely pleasant equality of $\mathcal O_X(U)$-modules:

$$(\mathcal F \otimes_{ \mathcal O_X} \mathcal G)(U)= \mathcal F(U) \otimes_{\mathcal O_X(U)} \mathcal G(U) $$

However if $U$ is not affine all hell can break loose:
For example if $X=\mathbb P^1_\mathbb C$ is the complex projective line and if $\mathcal F=\mathcal O_X(1), \mathcal G=\mathcal O_X(-1)$, then for these quasi-coherent sheaves $\mathcal F \otimes_{ \mathcal O_X} \mathcal G=\mathcal O_X$, so that for $U=X$: $$(\mathcal F \otimes_{ \mathcal O_X} \mathcal G)(X)= \mathcal O_X(X) =\mathbb C \neq \mathcal F(X)\otimes _{\mathcal O_X(X)}\mathcal G(X)=\mathbb C^2\otimes_\mathbb C 0=0$$

Reference
The first displayed equality is proved in Qing-Liu, chapter 5, Proposition 1.12 (b), page 162.

Are you asking for the $\mathcal O_X(U)$ action on $(\mathcal F \otimes_{\mathcal O_X} \mathcal G)(U)$? If so, it works on the level of stalks. Let $f \in \mathcal O_X(U)$ and let $s \in (\mathcal F \otimes_{\mathcal O_X} \mathcal G)(U)$. $s$ can be thought of as a collection of compatible germs $s_p \in \mathcal F_p \otimes_{\mathcal O_{x, p}} \mathcal G_p$. Then $f \cdot s$ is the collection of germs $f_p \cdot s_p$, where the dot is the $\mathcal O_{x, p}$-module action on $\mathcal F_p \otimes_{\mathcal O_{x, p}} \mathcal G_p$. One just needs to verify that the collection of $f_ps_p$ is still a collection of compatible germs, which they are.