The Stacks project

Sorry, I don't see why this definition does not depend on the choice of the K-flat resolution. For two K-flat resolutions $K_1^{\bullet}$ and $K_2^{\bullet}$ of $F^{\bullet}$, is there always a quasi-isomorphism $K_1^{\bullet}\rightarrow K_2^{\bullet}$ (in order to use the Lemma 06YG above)?

No. To get around this the standard argument is as follows: if $K_1 \to M$ and $K_2 \to M$ are quasi-isomorphisms in $K(\mathcal{A})$ where $\mathcal{A}$ is any abelian category, then there exists an object $N$ of $K(\mathcal{A})$ and quasi-isomorphisms $N \to K_1$ and $N \to K_2$ making the square commute. This follows from the fact that the collection of all quasi-isomorphisms forms a saturated multiplicative system of arrows compatible with the triangulated category structure, see Lemma 13.11.2 (and click through to the lemmas it relies on for more information). Then finally, in your case, you can choose a quasi-isomorphism $K \to N$ with $K$ K-flat to get $K \to K_1$ and $K \to K_2$ to which Lemma 20.26.13 applies.

When the map $K_2 \to M$ is termwise surjective, you can take $N = K_1 \times_M K_2$ and proceed as before (without appealing to the general results on the triangulated structure of the homotopy category). Note that the existence in Lemma 20.26.12 does produce such maps and that this is enough for our purposes.

i will add this discussion, if others chime in.