I want to show the following. Suppose $X$ is a smooth manifold and F,G are sheaves of $C^{\infty}_{X}$-modules, then the natural map $Hom(F,G)\to Hom(F(X),G(X))$ is injective.
It's easy to see that if $\phi_x:F(x)\to G(x)$ is the constant map to identity, then by the compatibility condition $\forall q \in F(U), q=s_{U}, s\in F(X), U$ is open in $X$, $q$ is mapped to identity by $\phi_u$. But I don't know how to show generally all $q$ in $F(U)$ are mapped to the identity.
$C^{\infty}$-sheaves are soft sheaves: Any section over a closed subset $V \subset X$ extends to a section over $X$. By definition a section over a closed subset $V$ is represented by a section over a suitable open neighbourhood of $V$.
Proof of the original claim: To show that $\phi: \mathcal F \longrightarrow \mathcal G$ is zero as morphism of sheaves, it suffices to prove that for every $x \in X$ the morphism of stalks $\phi_x: \mathcal F_x \longrightarrow \mathcal G_x$ is zero. Because the sheaf $\mathcal F$ is soft, any germ $f_x \in \mathcal F_x$ extends to a global section $f \in \mathcal F(X)$, which maps to $\phi_X(f)=0\in \mathcal G(X)$. From the canonical commutativity we get $\phi_x(f_x) = 0 \in \mathcal G_x$, q.e.d.
Note. I have corrected the original proof because $C^{\infty}$-sheaves are not flabby, as Georges pointed out.