Let $S$ be a scheme and $\mathcal{E}$ a vector bundle over $S$ of rank n. Given a global section of $\mathcal{E}$ we define $$V(s)=\{p\in S\mid\operatorname{germ}_ps\in\mathfrak{m}_p\mathcal{O}^n_{S,p}\}$$
This is a closed subscheme of $S$. Given a scheme morphism $(f,f^{\sharp}): X \to S$ such that $f^{*}(s)=0$ we can see that, $$\operatorname{germ}_{f(x)}s \otimes 1= 0 \in \mathcal{O}_{S,f(x)} \otimes_{\mathcal{O}_{S,f(x)}} \mathcal{O}_{X,x} $$
Since $\mathcal{E}$ is locally free we see that, $$\mathcal{O}_{S,f(x)} \otimes_{\mathcal{O}_{S,x}} \mathcal{O}_{X,x} \cong \mathcal{O}^n_{X,x}$$ and we can identify $\operatorname{germ}_{f(x)}s$ with a tuple $(s_1,\dots,s_n)\in \mathcal{O}^n_{S,f(x)}$ such that $f^{\sharp}_{f(x)}(s_i)=0$. Now we have that $\operatorname{Ker}\left(f^{\sharp}_{f(x)}\right) \subset \mathfrak{m}_{f(x)}$ and the claim is true.
I am trying to prove the converse. I don't know how can I deduce from the fact $f^{\sharp}_{f(x)}(s_i) \in \mathfrak{m}_{x},$ that $f^{*}(s)=0$.