Let $\mathcal{S}^{q}$ be the presheaf of singular q-cochains on a topological space $X$, that is the functor $\mathcal{S}^{q} \colon \mathcal{O}(X)^{op} \to \mathbb{Z}$-$\mathsf{Mod}$ which is given by $\mathcal{S}^{q}(U)= \operatorname{Hom}(S_{q}(U), \mathbb{Z})$, where $S_q(U)$ is the abelian group of singular $q$-chains on $X$. I understood from my previous question why $\mathcal{S}^q$ is not a sheaf but a presheaf.
Now if I take an arbitrary singular chain $\sigma \colon \Delta^q \to X$, the image $\sigma(\Delta^q)$ lies in some open set $W \subset X.$Is the germ of a cochain $f \in \mathcal{S}^q(W)$ satisfying $\langle f, \sigma \rangle =1$ equal to $0$ at any point $x \in W$? I feel if it is true, all of the germs of the presheaves of singular cochains must be $0$ except for the following example as we can chose an open covering $\{U, V \}$ such that $W \not \subset U$ and, $W \not \subset V$.
I found a example of non-zero germ of singular cochain $f \in \mathcal{S}^q(W)$ at $x \in W$ satisfying $\langle f, \sigma \rangle=1$ which image $\sigma(\Delta^q)$ is equal to a point $x \in X$. I'm still confused that the element of sections of the sheafication of the presheaf $\mathcal{S}^q$ are consisted only such elements?