What does the derived functor send roofs of one derived category to? For reference, I am working out of Pramod Achar's notes here, and my confusion centres around definition 7 on page 15.
For $F: \mathcal A \to \mathcal B$ a left exact functor, and $\mathcal R$ an adapted class for $F$ (ie $F$ is exact on $\mathcal R$ and $\mathcal R$ is large enough in $\mathcal A$), the derived functor $$ RF: D^+(\mathcal A) \to D^+(\mathcal B)$$ is defined by sending $A^\bullet$ to the complex $F(R^\bullet)$, where $R^\bullet$ is an $\mathcal R$-resolution of $A^\bullet$ (a quasi-isomorphism $t : A^\bullet \to R^\bullet$ exists).
I am hoping to understand the corresponding morphism of $\text{Hom}_{D^+(\mathcal A)}(A^\bullet, B^\bullet)$ is under $RF$.
I know that a morphism in $\text{Hom}_{D^+(\mathcal A)}(A^\bullet, B^\bullet)$ is a roof $$A^\bullet \xleftarrow{q} X^\bullet \rightarrow B^\bullet$$ where $q$ is qis, so my guess would be that (with $s: B^\bullet \to Q^\bullet$ an $\mathcal R$-resolution of $B^\bullet$) the roof is sent to $$F(R^\bullet) \xleftarrow{F(t \circ q)} F(X^\bullet) \rightarrow F(Q^\bullet).$$
However I don't think this is very well-defined, unless I add the constraint that the $X^\bullet$ is also in $\mathcal R$ (if this were the case, then the fact that $F$ is exact on $\mathcal R$ tells me that $F(t \circ q)$ is qis, from which I get an equivalence class for the roof in $D^+(\mathcal B)$).
Since in the definitions of the derived categories I don't get such a constraint, my suspicion is that I am slightly wrong here, but I can't think of how to fix this. I have tried further unpacking the equivalence classes on the roofs of $D^+(\mathcal B)$ to show that we don't need to take $F(X^\bullet)$, but I haven't managed to show that sending the roof $A^\bullet \xleftarrow{q} X^\bullet \rightarrow B^\bullet$ to any equivalence class $F(R^\bullet) \leftarrow Y^\bullet \rightarrow F(Q^\bullet)$ is well-defined, at least functorially-speaking.
 A: First of all you can use the roof in the opposite direction $A^\bullet \rightarrow X^\bullet \xleftarrow{q} B^\bullet$ where $q$ is a quasi isomorphism. now you have a quasi-isomorphism from $X\to R$ so you get an equivalent  roof such that the middle term is in $R$ and basically your idea work. but to made it precise you have two choice:
you can consider the subcategory $\mathcal R\to \mathcal A$ and use the previous paragraph to show that $S^{-1}\mathcal R\to S^{-1}\mathcal A$ is an equivalence of categories(where $S$ is the set of quasi isomorphism). then as you said it is easy to define a map $S^{-1}\mathcal{R}\to \mathcal B$ which is enough fo most practical purposes.
or you can first choose a complex in $\mathcal R$ which is quasi-isomorphism to each element of $\mathcal A$ and then using this choices define the $RF$ it is easy to check that this map is well defined when you work with only one functor $F$ and every two choices give equivalent functors. of course, when you want to talk about the combinacation of two different derived functor things become uglier.
Or you can use the language of higher category theory. any way the point is that you can make these things precise so usually people just accept this things in practice.
