Intuition for the satellite of a functor

I don't know if this can be considered as "intuition", anyway

another way to think about derived functors is the following: given an abelian category $mathcal{A}$, you can define its derived category $mathcal{D}(mathcal{A})$ (or its variants of bounded complexes in one or both directions). You get $mathcal{D}(mathcal{A})$ by "localizing" the homotopy category of complexes $mathcal{K}(mathcal{A})$, where the objects are complexes in $mathcal{A}$ and maps are maps of complexes up to homotopy, at the system of quasi-isomorphisms.

There is a natural localization functor $pi_A:mathcal{K}(mathcal{A})to mathcal{D}(mathcal{A})$. If $F:mathcal{A}to mathcal{B}$ is your additive functor between abelian categories, and $mathcal{K}(F):mathcal{K}(mathcal{A})to mathcal{K}(mathcal{B})$ is the induced functor, it is natural to ask for an "extension" of $mathcal{K}(F)$ to the derived category $mathcal{D}(mathcal{A})$, with values in $mathcal{D}(mathcal{B})$. In other words this would be a functor $RF:mathcal{D}(mathcal{A})to mathcal{D}(mathcal{B})$, such that $pi_B circ mathcal{K}(F) = RF circ pi_A$.

This is not possible to find in general, and the problem is that $mathcal{K}(F)$ may not send quasi-isomorphisms into quasi-isomorphisms. The best you can ask for is a functor $RF:mathcal{D}(mathcal{A})to mathcal{D}(mathcal{B})$, with a natural transformation $eta:pi_B circ mathcal{K}(F) to RFcirc pi_A$ having a universal property among such functors and natural transformations (this is a particular case of a Kan extension).

Such an $RF$ (unique up to isomorphism) is called the (total) right derived functor of $F$. One way to think about it is as "the" functor between the derived categories which approximates $mathcal{K}(F)$ in the best possible way. You can recover the single derived functors $R^iF$ by taking the cohomology objects of $RF$. In most cases the derived functor is constructed by using resolutions, by injective (or projective, if you're defining left derived functors) objects, or more generally by suitable subcategories of $mathcal{A}$.

In your case, if you also assume that $F$ is right exact, then the satellite functors coincide with the (left) derived ones, and so I guess that it follows that they can be calculated by the formulas you wrote.

The idea of the "best approximation" of $mathcal{K}(F)$ on the derived category seems very natural to me, and a satisfactory answer to the question "why derived functors". If you were asking specifically about satellites, then I don't know.