Let $ \mathcal{F }$ be a sheaf of abelian groups on a topological space $ X $. Let $ s_{i} \in \mathcal{F} ( U _ { i } ) $ be sections on open sets $ U _{i} $ (which do not necessarily form a cover of $ X $). Let $ \mathcal{G} $ be the sheaf generated by these sections i.e. the intersection of all sub-sheaves of $ \mathcal{F }$ that contain these local sections. Is it true that if $ V \subset X $ is open such that $ V \not \subset \bigcup_{i} U_{i} $, then $ \mathcal{G} ( V) = 0 $?
EDIT: This is false, as shown by the counterexample below (extend the function with compact support in $ (-1,1)$ by zero). However, what is true is the following: if $ V $ does not intersect the union of $ U_{i} $, then $ G(V) $ is $ 0 $ by looking at the stalks of the sub-sheaf given by extension of $ \mathcal{F}$ by zero outside the union of $ U_{i} $.
The answer is yes, because for an open subset $j:U \hookrightarrow X$, we have a natural injection $j_!j^{-1}\mathcal F \hookrightarrow \mathcal F$.
In your case, take $U$ to be the union of the $U_i$. Then $j_!j^{-1}\mathcal F$ contains all $s_i$ and is zero on any open set, that is not contained in $U$ (By the definition of the $j_!$-functor, see Exercise 1.19 in Hartshorne).