In this short paper by G. Maltsiniotis derived functors are presented as Kan extensions along the localization functor.
I began studying derived categories only a couple of months ago, so I'm not at all an expert, but this approach is completely new to me, as I've not found it in any of the book in my hand (the most advanced is Categories and Sheaves but I've never read it sistematically, so maybe I haven't noticed it there).
Can you give, if it exists, a reference which rewrites some of the basics on derived functors taking into account the "Kan extensions perspective"?
In the context of model categories, the left (reps. right) derived functors are a generalization of the traditional notion of a derived functor. The left (resp. right) derived functor $LF$ of a functor $F\colon \mathbf C\to \mathbf D$ where $\mathbf C$ is a model category is exactly a right (resp. left) Kan extension along the localization $\gamma\colon \mathbf C\to\operatorname{Ho}(\mathbf C)$. This terminology of derived functors stems from what happens in the category of chain complexes. In general, computing $LF(X)$ is computing $F(QX)$ where $QX$ is a cofibrant replacement of $X$ where we view an $R$-module $X$ as a chain complex concentrated in degree 0 (the definition of $LF$ is unique up to natural isomorphism). In the category of chain complexes, $QX$ is a projective resolution of $X$, and hence the homology of $LF(X)$ for a right exact functor $F$ is the left derived functors of $F$. For example, if $F$ is $-\otimes_R A$, then the homology of $LF$ will be $\operatorname{Tor}_*^R(-,A)$.
We have a similar story for right derived functors (which are left Kan extensions along the localization functor). Here, $RF(X)=F(RX)$ where $RX$ is a fibrant replacement. For chain complexes, this is an injective resolution, and so the cohomology of $RF(X)$ is the right derived functors of $F$.
If you are unfamiliar with model categories, I'd recommend Homotopy theories and model categories by Dwyer and Spalinski. In particular, example 9.6 deals with what I just outlined.