restrictions of rational sections of an invertible sheaf. Let $(X,\mathscr O_X)$ be an integral scheme with function field $K(X)$ and let $\mathscr L$ be an invertible sheaf on $X$. Moreover suppose that $\{U_i\}$ is an open covering of $X$ such that $\mathscr L|_{U_i}\cong\mathscr O_X|_{U_i}$.
Fix a non-zero rational section $s\in H^0(X,\mathscr L\otimes_{\mathscr O_X} K(X))$, then why $s|_{U_i}\neq 0$ for any $i$?
In practice on the open sets where $\mathscr L$ is locally trivial $s$ can't be restricted to $0$.
 A: With the same amount of effort, we get a much more general statement:
Let $\mathcal F$ be a torsion free sheaf of modules on an integral scheme $X$. Then $\mathcal F_x=0$ for one single $x \in X$ already implies $\mathcal F=0$. In particular $\mathcal F_{|U}=0$ for one single non-empty open (not necessarily affine) $U \subset X$ implies $\mathcal F=0$.
Proof: Choose some affine open $\operatorname{Spec}A=U \subset X$ which contains $x$. On $U$, $\mathcal F_{|U}$ corresponds to some torsion-free $A$-module $M$ and $\mathcal F_x=0$ means $M_{\mathfrak p}=0$ for some prime of $A$. But the localization of a torsion-free non-zero module over an integral domain is never zero, hence we deduce $M=0$ and thus $\mathcal F_{|U}=0$.
In particluar $\mathcal F_{\eta}=0$ for the generic point $\eta \in X$. Now take some affine open cover $X = \bigcup U_i$ and repeat the argument above for any $U_i$ with $\eta$ in the role of $x$ (Note that $\eta$ is contained in any non-empty open set). We deduce $\mathcal F_{|U_i}=0$ for any $i$, i.e. $\mathcal F=0$.

In your case, let $\mathcal F$ be the subsheaf of $\mathcal L \otimes K(X)$ generated by $s$, i.e. the image of the morphism $\mathcal O_X \to \mathcal L \otimes K(X)$ given by $s$.
A: Over a trivialization, the rational section is given by elements of the function field which is globally glued using the transition maps of the invertible sheaf $\mathcal L$. Therefore, if it restricts to zero over one open set, then it is zero everywhere.
