Relations between $R^fG$ and either $\mathbb{C}^fG$ or $\mathbb{Z}^fG$. Denote by $RG$ the group ring of the group $G$ over the commutative ring $R$. A result by Passman saying that if $R$ is a commutative ring then
$$RG=R\otimes_{\mathbb{Z}}\mathbb{Z}G.$$
As a result, the ring structure of $\mathbb{Z}G$ contains more data on the structure of the group $G$ then any other group ring over $G$.
It is preaty clear to me that there will not be such a nice equation in the twisted case, however I was wondering if we can still someting in the spirit of the above.
More explicitly, let $f\in H^2(G,R^*)$ ($f$ is a $2$-cohomological class), can we say something about the ring structure of the twisted group ring $R^fG$, from a corresponding twisted group ring (I am looking for such correspondence), $\mathbb{C}^{\tilde{f}}G$ or $\mathbb{Z}^{\bar{f}}G$.
Here $\tilde{f}\in H^2(G,\mathbb{C}^*)$ and $\bar{f}\in H^2(G,\mathbb{Z}^*)$ are the cohomological classes corresponding to $f$.
So far, the best relation that I found comes from the universal cofficiant theorem that say that the following sequance is exact
$$ 0\rightarrow Ext(G/G^{\shortmid},R^*)\rightarrow H^2(G,R^*)\rightarrow Hom(H^2(G,\mathbb{C}^*),R^*)\rightarrow 0.$$
 A: This is not a full answer, but maybe it can helps...
As mt_ said, $\overline{f}$ and $\tilde{f}$ does not mean anything here. However, there is a map $H^2(G,\mathbb{Z}^\times)\rightarrow H^2(G,R^\times)$ (at least if $G$ acts trivialy on $-1\in R^\times$). So that if $f\in H^2(G,\mathbb{Z}^\times)$ there is a corresponding element in $f_R\in H^2(G,R^\times)$. In that case, it is trivial to see that there is an isomorphism
$$R\otimes \mathbb{Z}^f G\overset{\sim}\longrightarrow R^{f_R}G$$
(I have never heard of this Passman result, unless I am misunderstanding something, the isomorphism $RG=R\otimes \mathbb{Z}G$ is trivial).
The case of the usual (non twisted) group rings corresponds to $f=0\in H^2(G,\mathbb{Z}^\times)$ (that is to the 2-cocycle $G\times G\rightarrow \{\pm1\}$ which is constant equals to $1$).
However, the map $H^2(G,\mathbb{Z}^\times)\rightarrow H^2(G,R^\times)$  is usually not onto, so that there may be twisted group rings that do not come from $\mathbb{Z}$.
I am not sure to understand fully what do you expect with $\mathbb{C}^*$.
