${\rm Ext}\ ({\bf Z}_n,G) =G/nG$ so that if $G={\bf Q},\ {\bf R}$ then ${\rm Ext}\ ({\bf Z}_n,G)=0$
Hence UCT implies $$ H^n(C;G) = {\rm Hom}_{\bf Z} ( H_n(C);G) $$
Hence $$ H^n(C;G) =H_n(C)/{\rm Tor}(H_n(C)) \ \otimes_{\bf Z} G$$
Why we do consider $H^n(C;{\bf R})$ ? $H^n(C;{\bf Q})$ is not enough ?
If I understand you correctly, you are asking why we consider $H^n(C; \mathbb{R})$ at all, considering that it is determined by $H^n(C; \mathbb{Q}) \otimes_\mathbb{Q} \mathbb{R}$. Indeed these are vector spaces, and the only invariant is dimension, aka the Betti numbers.
One reason I can think of is that real cohomology (say of manifolds) can be instead computed by de Rham cohomology, which is sometimes easier to manage.
In the other direction, for rational cohomology, you can construct a model for the rational homotopy of a space $X$, that is a CDGA $A_X$ which is minimal and whose cohomology is isomorphic to $H^*(X; \mathbb{Q})$. But more than that, you can also retrieve rational homotopy groups $\pi_*(X) \otimes \mathbb{Q}$ from $A_X$. As far as I understand, the theory is more interesting than for real homotopy (where de Rham forms are a real model), because the models $A_{PL}(X)$ exist for arbitrary topological spaces, whereas $A_{dR}(X)$ is only defined for manifolds. See for example Algebraic models in geometry by Félix, Oprea, Tanré.