I am very new in studying algebraic topology and cohomology. $C^p(K;G)=Hom(C_p(K),G)$ are cochains of $K$, $Z^p(K;G)$ are cocycles, i.e. kernels of delta homomorphisms, $B^p(K;G)$ are coboundaries of delta homomorphisms and $H^p(K;G)=Z^p(K;G)/B^p(K;G)$ are cohomology groups of $K$ with coefficients in $G$. [Munkres, Elements of algebraic topology, chapter 42]
Since we can easily see that if A is free abelian group of finite rank n, then Hom(A,G) is isomorphic to $G\bigoplus G\bigoplus ... \bigoplus G$ (n summands), one can conlude that $Hom(A,\mathbb{Z})$ is isomorphic to n summands $\mathbb{Z}$, so it is a free abelian group of rank n, when A is free abelian group of rank n. So, when we take K to be a finite complex, cochains are free abelian groups of finite rank, which means we can repeat the story of homology groups?
Am I right?
Yes, this is correct. More generally, the same conclusion holds as long as $G$ is finitely generated. More surprisingly, this conclusion actually doesn't depend on $K$ being finite: it only depends on the homology of $K$ being finitely generated. Proving this requires a much more careful analysis of what happens to homology when you apply the functor $\operatorname{Hom}(-,G)$, known as the universal coefficient theorem.