I'll reproduce part of Hartshorne's exercise II.1.19 (b) (the rest is not important to the question):
Let $Z$ be a closed subset of a topological space $X$ an $i:Z\to X$ the inclusion. If $\mathcal{F}$ is a sheaf on $U:=X\setminus Z$, let $j_!\mathcal{F}$ be the sheaf on $X$ associated to the presheaf $V\mapsto \mathcal{F}(V)$ if $V\subset U$ and $V\mapsto 0$ otherwise. [...]
Let $\mathcal{G}$ be the presheaf defined above. Suppose there is an open $V\subset X$ such that $V\supsetneq U$. By definition of $\mathcal{G}$, we have $\mathcal{G}(V)=0$. Consequently, applying the restriction we find that $\mathcal{G}(V\cap U)=0$, therefore $\mathcal{F}(U)=\mathcal{G}(U)=\mathcal{G}(V\cap U)=0$. Which ultimately means $\mathcal{F}=0$.
This is very weird. Am I missing something?
The presheaf $\mathcal{G}$ has $$ \mathcal{G}(V)= \begin{cases} 0&V\cap Z\ne \varnothing\\ \mathcal{F}(V)&V\subseteq U. \end{cases} $$ I understand your concern as being that given $V$ strictly containing $U$ (such as $X$), we get that $\mathcal{G}(V)=0$. Then, you wish to apply the restriction morphism $\mathcal{G}(V)\to \mathcal{G}(V\cap U)=\mathcal{G}(U)$ to deduce that $\mathcal{G}(U)=0$, which ought not be true. I think you are assuming that the restriction map need be surjective, which is definitely not the case.
Indeed, just look at the sheaf of holomorphic functions on $\mathbb{P}^1$. Indeed, $\mathcal{O}(\Bbb{P}^1)\to \mathcal{O}(U)$ is not surjective in general, as the former contains only constant functions.