The idea is that a cochain $\varphi \in C^n(X)$ is compactly supported if there's a $K \subseteq X$ compact subset of $X$ such that $\varphi|_{C_n(X \setminus K)} = 0$.
Edit a little remark: for every $K$ compact subset of $X$ there's an embedding $i \colon X \setminus K \hookrightarrow X$ which give rise to a injective embedding of chain complexes $i_* \colon C_\bullet(X \setminus K) \to C_\bullet (X)$, so we can think of $C_n(X \setminus K)$ as being a submodule of $C_n(X)$ and to be exact what I meant above by $\varphi|_{C_(X \setminus K)}$ should be written more formally as $\varphi|_{i_*(C_n(X \setminus K))}$.
So compactly supported co-chain of $X$ are those co-chains in $C^\bullet(X)$ that vanish on all the simplexes that have image contained in a subspace $X \setminus K$ (for some $K$ compact subset of $X$), i.e. those simplexes $\sigma \colon \Delta^n \to X$ that factors through the inclusion map $i \colon X \setminus K \to X$.
You can find out more about this in Hatcher's book Algebraic Topology.
Let $X$ be an abstract simplicial complex and $\lvert X \rvert$ its geometric realization which is a topological space representable as the union of geometric simplices. It is well-known that there exist natural isomorphisms $\phi_n^X : H_n(X) \to H_n(\lvert X \rvert)$ between the simplicial homology groups $H_n(X)$ of $X$ and the singular homology groups $H_n(\lvert X \rvert)$ of $\lvert X \rvert$. In particular, if $f : X\to Y$ is a simplicial map, then $\lvert f \rvert_* \circ \phi_n^X = \phi_n^Y \circ f_*$.
Using this, Qiaochu Yuan's comment answers your question. However, we can also see this without changing to topological spaces.
The simplicial chain complex of $X$ is based on oriented simplices. An orientation of an $n$-simplex $s^n = \{ v_0,\dots,v_n \}$ is an equivalence class $[v_{\pi(0)},\dots,v_{\pi(n)}]$, where $\pi$ ranges over all permutations of $\{ 0,\dots,n \}$. For $n > 0$ there exist two orientations, for $n = 0$ only one. The simplicial chain map $C_*(f) : C_*(X) \to C_*(Y)$ is given on the simplicial generators $\sigma^n = [v_0,\dots,v_n]$ of the free abelian group $C_n(X)$ by
$$C_n(f)([v_0,\dots,v_n]) = [f(v_0),\dots,f(v_n)].$$
In dimension $0$ we simply have
$$(*) \phantom{xx} C_0(f)(v) = f(v)$$
with the vertices $v$.
Now you consider a simplicial map $f : X \to X$ on a path connected $X$. In my understanding of your question you know that all vertices $v$ of $X$ (i.e. all simplicial generators of $C_0(X)$) are in the same homology class $g \in H_0(X)$ and that $H_0(X)$ is free abelian with $g$ as a generator. Let us call $g$ the canonical simplicial generator of $H_0(X)$. This provides a canonical isomorphism $H_0(X) \approx \mathbb{Z}$ (mapping $g$ to $1$). $H_0(X)$ of course has an alternative generator: This is $-g$.
But now it obvious from $(*)$ that $f_*(g) = g$ since both $v$ and $f(v)$ are vertices and therefore in the same homology class.
Best Answer
$H^0(X; G)$ is the dual of $H_0(X; G) = \bigoplus_\alpha G$, hence $\prod_\alpha G$. This can be seen by explicitly computing the cocycles as you indicated, which are $G$-valued $0$-cochains that are constant on the path components of $X$, i.e., constant on $X_\alpha$. This is the direct product of the groups $G$ indexed over the path components. There are no nontrivial $0$-coboundaries because of degree reasons.