Many thanks ! (the terminology I was looking for is the Strong Approximation Theorem)

@reuns Thank you for your comment. Is there a result that guarentee me the existence of $f$? For instance, when $S$ is $C\setminus \{Q\}$ for a closed point $Q$?

@PhilTosteson Thank you very much. I agree with your comment. However, the question is rather then: how does one show that $f_X$ indeed comes from the functorial cone construction? (maybe I am missing something...)

After a small computation, it seems that $\operatorname{cone} h$ do depend on the differential of $C_{\bullet}$ and $C_{\bullet}'$ while $[\operatorname{coker}f\to \operatorname{coker} g]$ do not. Maybe this explains why $h$ cannot be arbitrary?

A priori $h$ is arbitrary here (but still such that it forms a morphism of distinguished triangles). Is this an issue regarding the truth of the statement?

Many thanks @LeoAlonso, it indeed seems to be the following lemma stacks.math.columbia.edu/tag/0CRS

Thank you @FernandoMuro for your reference. So what I actually understand from there is the following: if $f$ and $g$ are given, from the octahedral axiom there exists $h:C\to C'$ such that $(f,g,h)$ is a morphism of (distinguished) triangle. For this $h$, we further have that a distinguished triangle $C(f)\to C(g)\to C(h)\to [1]$ (and hence, if $f$ is a quasi-isomorphism, then $C(f)$ is acyclic and $C(g)\to C(h)$ is a quasi-isomorphism). But this is not true for all $h$. Is that right?