This (elementary and perfectly standard) example might help show the power of sheaves with non-constant coefficients:
First, think about the circle $S^1$. Suppose you want to understand (real) line bundles on the circle. You can certainly cover the circle with two open contractible subsets $U_1$ and $U_2$ (which you can take to be the complements of the north and south poles), and we know that any line bundle on a contractible space is trivial. So if you've got a line bundle $L$ over $S^1$, you can restrict it to either $U_i$ and get a trivial bundle $L_i$. $L$ is built from these $L_i$ and the way they they are patched together over $U_1\cap U_2$.
Now what does it mean to patch the $L_i$ together over $U_{12}=U_1\cap U_2$? It means choosing an isomorphism $L_1|U_{12}\rightarrow L_2|U_{12}$. For any $x\in U_{12}$, the restriction of this isomorphism to the fiber $L_x$ over $x$ is an isomorphism between 1-dimensional vector spaces, and so (after choosing bases) can be identified with an element of ${\bf R}^*$ (the non-zero reals). Therefore your patching consists of a continuous map
$$U_{12}\rightarrow {\mathbb R}^*$$
which is to say, a Cech 1-cocycle for the sheaf of continuous ${\bf R}^{*}$-valued functions.
Now of course you could build a line bundle in some other way, say by starting with two different contractible sets $U_1$ and $U_2$. When do two sets of patching data give isomorphic line bundles? A little thought reveals that the answer is: When and only when the corresponding cocycles give the same class in
$$H^1(S^1,G^{*})$$
with $ G^{*} $ being the sheaf of continuous ${\bf R}^*$-valued functions.
Therefore line bundles are classified by $H^1(S^1,G^{*})$. Now consider the exact sequence of sheaves
$$0 \rightarrow G \rightarrow G^*\rightarrow {\bf Z}/2{\bf Z}\rightarrow 0$$
where $G$ is the sheaf of continuous ${\bf R}$ valued functions, and the map on the left is exponentiation. Follow the long exact sequence of cohomology, use the fact that $G$ is acyclic, and conclude that $H^1(S^1,G^*)=H^1(S^1,{\bf Z}/2{\bf Z})={\bf Z}/2{\bf Z}$. In other words, there are exactly two real line bundles over $S^1$ --- and indeed there are: the cylinder and the Mobius strip.
Exercise: Do a similar calculation for ${\bf CP}^1$ (the Riemann sphere). Conclude that the set of (complex) line bundles is in one-one correspondence with $H^2({\bf CP}^1,{\bf Z})={\bf Z}$.