I feel this is somewhat naïve but it's always good to clear up confusion.
Let $X$ be a topological space, $Z$ a closed subspace, $U$ its complementary open, $i$ the inclusion of $Z$ into $X$ and $j$ the inclusion of $U$ into $X$. For a sheaf $\mathcal{F}$ on $X$, there is a well-known short exact sequence
$$0\to j_!(\mathcal{F}|_U)\to \mathcal{F}\to i_*(\mathcal{F}|_Z)\to 0.$$
This doesn't split in general, so what is wrong with the following argument? We can define a surjective map of sheaves $\mathcal{F} \to j_!(\mathcal{F}|_U)$ by defining a map on the stalks which is an isomorphism over $U$ and zero otherwise. The kernel of this map is a sheaf on $X$ supported on $Z$, whose stalks are isomorphic to those of $X$ over $Z$ and zero otherwise, which implies that the kernel is $i_*(\mathcal{F}|_Z)$.
Thanks in advance!
The map $\mathcal{F}\to j_!(\mathcal{F}|_U)$ you defined is not in general a map of sheaves, because it does not commute with restriction. Suppose one has a section $f\in \mathcal{F}(V)$ where $V$ is an open subset which intersects $Z$, and also has $f|_{U\cap V}\neq 0$: then the image of $f$ in $j_!(\mathcal{F}|_U)(V)$ is zero, so the restriction of that image to $j_!(\mathcal{F}|_U)(U\cap V)=\mathcal{F}(U\cap V)$ should be zero. But the restriction of $f$ to $U\cap V$ is nonzero by assumption.