I am trying to understand why the compactly supported sections of a vector bundle form a cosheaf. I have proven that they form a precosheaf: simply extending the compactly supported section by zero proves this. However, checking the local to global property of a cosheaf I do not manage to prove. I found an argument only in Costello-Gwilliam: Factorization algebras in Quantum Field Theory, but I don't understand it. How can one use a partition of unity, to check the exactness of the following sequence: \begin{equation*} \bigoplus_{(i, j)} F\left(U_i \cap U_j\right) \stackrel{\beta}{\longrightarrow} \bigoplus_i F\left(U_i\right) \stackrel{\alpha}{\longrightarrow} F(U) \longrightarrow 0 \end{equation*} ,where $F$ is the functor, which sends open sets to compactly supported sections of a vector bundle and $\alpha=\sum_{i \in I} \limits i_{U,U_i}$ and $\beta=\sum_{(i,j)} \limits(i_{U_i,U_i \cap U_j} -i_{U_j,U_i \cap U_j})$ and $i_{A,B}$ is the linear map coming from the inclusion $A\to B$ under the functor $F$,namely $i_{A,B}=F(A)\to F(B)$?
EDIT: To be more specific, my concrete confusion comes in how to prove $\text{im}(\beta)=\text{ker}(\alpha)$. Does one need partitions of unity for that? To show that alpha is surjective, which is part of the exactness, one does have to use a partition of unity, and as A. Thomas Yerger noted, it follows more or less by definition.
Here's a tentative answer under the assumption that $\beta$ is defined incorrectly, slightly. I think it should be as follows. Let $F$ be a vector bundle, and think of it as a functor taking values in vector spaces of sections. Then we have $\beta: \bigoplus_{i,j} F(U_i \cap U_j) \to \bigoplus_{i} F(U_i)$ and be defined by sending $$(\phi: U_i \cap U_j \to F) \mapsto \left( (i_*\phi:U_i \to F), (-i_*\phi:U_j \to F) \right)$$, where $i_*$ is the extension by $0$ of the inclusion, as in the question.
In this form, we consider the kernel of $\alpha$. This consists of all tuples of sections $\varphi_i: U_i \to F$ such that the $U_i$ sum to $0$ when you extend them all by $0$ to sections over all of $U$. For example, this happens when you have sections which are $0$ on the complement $U - \bigcup_{i,j} (U_i \cap U_j)$ and have opposite signs on the intersections, so that their sum is $0$. Well, this is exactly what $\beta$'s image is, by construction. We are sending things on the intersection to the same place with opposite sign so that they can be cancelled by $\alpha$.