This is actually a standard property of the homotopy category of complexes on which construction of the derived category is based, formulated in unusual way.
For a quasi-isomorphism $s:X'\to X$ and $f\in Hom(X',Y)$ denote by $fs^{-1}$ the image of $f$ in the first colimit(in a second we will justify the choice of notation). By description of the filtered colimit, two images $fs^{-1}$ and $gt^{-1}$(for $t:X''\to X,g\in Hom(X'',Y)$) coincide iff there exists $X'''$ and qisms $p:X'''\to X',q:X'''\to X''$ and a map $r\in Hom(X''',Y)$ such that $sp=tq,fp=gq$. This is exactly the condition for fractions $fs^{-1}$ and $gt^{-1}$ to be equivalent in the right localization by qisms(see e.g. Weibel, 10.3).
Analogously, the third colimit can be identified with the set of equivalence classes of left fractions. Now, Ore conditions guarantee that these two sets of equivalence classes coincide, which is precisely the isomorphism you are looking for.
Edit: As Denis-Charles Cisinski explains, the answer to B is negative. Indeed, let's pick a non-zero object $A$ in $C$ and consider, for instance, $X=A[0], Y=A\xrightarrow{1} A$ where $Y$ is a contractible complex concentrated in degrees $-1,0$. The second and third colimits are zero as the category of quasi-isomorphisms $Y\to Y'$ has a final object $Y\to 0$. But the first colimit is not zero: for $X'=X$ there is a non-zero element $Id_A\in Hom(A,A)=Hom^0(X,Y)$ which survives in the colimit because for a quasi-isomorphism $X'\to X$ the map $(X')^0\to A$ has to be non-zero.