Let $E$ be a globally generated vector bundle of rank $r$. Let $V$ be a subspace of $H^0(S,E)$ of dimension $r$. We have the evaluation map $ev:V\otimes \mathcal{O}_S\longrightarrow E$. Why is this map injective?
If $H^0(S,E)$ has as basis $s_1,\cdots, s_n$ and $V$ has as basis $s_1,\cdots,s_r$, then $ev$ sends the standard basis $e_i$ to $s_i$, for $i=i,\cdots,r$. At the stalk level, it is map between two free modules of rank r, $O_x\longrightarrow E_x$, sending ${(e_i)}_x$ to ${(s_i)}_x$, again $i:i,\cdots,r$. But these $(s_i)_x$ may not be a basis for $E_x$. So why is it injective?
This is not true. Take, for instance, $E = O(1) \oplus O(1)$ on $P^1$ and the space $V$ to be the global sections of the first summand. Then the evaluation map $V\otimes O \to E$ factors as $$ V \otimes O \to O(1) \hookrightarrow O(1) \oplus O(1) = E. $$ So, its kernel is $O(-1)$ and it is not injective.