The problem is that mapping spaces are not automatically functorial (essentially because you don't have a strict composition in quasicategories), so the usual formula for the represented presheaf doesn't define a functor. The construction of a "mapping space" functor and everything obtained from it, like the Yoneda embedding, is one of the main applications of straightening.

That's exactly what the machinery of faithfully flat descent does for you. It tells you precisely what additional structure you need to endow your $C_*$ with in order to find a $D_*$ as desired. I don't think you can turn this into a manageable list of obstructions in general, but if your complex is of finite length, this in principle turns into a finite amount of coherences.

They always differ for nonabelian $\pi_1$, but for an easy example just think of a perfect group.

Unpointed homotopy classes $[S^1,Y]$ are given by the set of conjugacy classes in $\pi_1(Y)$, not $H_1(Y)$. (The former is the orbits of the conjugacy action of $\pi_1$ on itself in the category of sets, the latter in the category of groups (i.e. the abelianisation.)

The polynomial equation in question is of degree 16 (given by the 17th cyclotomic polynomial).

As written, this is false. What follows from Whitehead is that given a map $f: X\to Y$ between connected spaces, which induces an equivalence on loop spaces, it was an equivalence to begin with. This follows for example also from the fact that loop space and the bar construction constitute inverse equivalences between connected spaces and grouplike $\mathbb{E} _1$-spaces.

There is a map of spectra $\mathbb{S}\to K(\mathbb{Z})$. I'm not completely sure, but I'd expect it to be injective on $\pi_3$.

Just write "there is a computable function $d(f)$ such that for all $f$..."

In Tyler's example, can't we just observe that for any finite simplicial set $K$, only finitely many maps on $\pi_1$ are induced by endomorphisms $K\to K$? This is because there are only finitely many maps $K\to K$ in total.

Yes it is. He explained how the trivial solution $(a, b) =(0, n) $, which was excluded in the question, gives rise to a solution in positive integers $(a, b)=(n^3,n)$ for any $n$.

Ah, my bad, I didn't see the condition that the classes are cohomologous.