$\newcommand{\hy}{\textrm{-}}$ Let $\mathcal{R}_X$ be a sheaf of rings on a smooth manifold $X$, let $\mathcal{R}_X\hy \mathsf{Mod}$ be the abelian category of $\mathcal{R}_X$-modules and let $D(X)=D(\mathcal{R}_X\hy \mathsf{Mod})$ be its derived category.
Assume that $\mathcal{C}^{\bullet}$ is a complex of $\mathcal{R}_X$-modules on $X$ and let $f: X\to Y$ be a continuous map.
Clearly, $f_*\mathcal{C}^{\bullet}$ is a complex of $\mathcal{R}_Y$-modules but it is not an element of the derived category $D(Y)$ yet. As far as I understand, the total derived functor $Rf_*\mathcal{C}^{\bullet}$ is the embedding of $f_*\mathcal{C}^{\bullet}$ in $D(Y)$.
In light of the above, I would like to know if it makes sense to talk about the hypercohomology $\mathbb H^{\bullet}(Y, f_*\mathcal{C}^{\bullet})$ of $f_*\mathcal{C}^{\bullet}$ at all because I have never seen such an object defined in the literature. In the literature, I have only seen $\mathbb H^{\bullet}(Y, Rf_*\mathcal{C}^{\bullet})$. This is very confusing because on the other hand, for an $\mathcal{R}_X$-module $\mathcal{M}$, I have seen in the literature the sheaf cohomology $H^{\bullet}(Y, f_*M)$ defined. Can someone elaborate on that?
The functor $f_*$ does extend to a functor between categories of chain complexes and so it makes perfect sense to talk about the cohomology of $f_*\mathcal{C}^{\cdot}$.
Be careful though, while we can view the object $f_*\mathcal{C}^{\cdot}$ as an element of the derived category (just apply the functor from chain complexes to the derived category), the functor $f_*$ itself is in general only left exact and therefore does not descend to a functor between derived categories. Typically people want constructions that are functorial, hence the emphasis on derived functors.