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 $\eta:\pi_B \circ \mathcal{K}(F) \to RF\circ \pi_A$ 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.
It seems this question was considered in:
William W Adams, Marc A Rieffel.
Adjoint functors and derived functors with an application to the cohomology of semigroups
Journal of Algebra, V. 7, N 1, 1967, 25-34
Best Answer
This is far from being a technical issue, there are many examples when it fails. Suppose that A is the category of $\mathbb F_p$-vector spaces, $B = C$ the category of abelian groups, $F$ the embedding, $G = \mathrm{Hom}(\mathbb Z/p\mathbb Z, -)$. Then it is easy to see that $DF = F$, $D(GF) = GF$, but $DG\circ F \neq G \circ F$, and the equality does not hold.