On page 193 of Hatcher at the bottom. He says
$$0 \to B_{n-1} \to Z_{n-1} \to H_{n-1}(C) \to 0 (vi)$$
Note that the group $Coker(i_{n-1}^*)$ that we are interested in is $H^1(F;G)$ where $F$ is the free resolution in (vi). Part (b) of the following lemma therefore shows that $Coker i∗_{n−1}$ depends only on $H_{n−1}(C)$ and $G$.
Which two resolution is he using to derive this result?
Is it $(iv)$ and $(v)$?? Very confused.
How does he pass through $H^n(C;G)$?
Math: the claim is that if we change the chain complex $C$ and pick a totally different complex C', requiring only that it have the same homology in the $n-1$ spot, then the complex $\text{Hom}(C, G)$ and the complex $\text{Hom}(C', G)$ would still have the same $\text{cok}(\iota^*_{n-1})$ (up to canonical isomorphism). The claim is attained by taking the sequence (vi) and the analogous sequence (vi') for $C'$ and applying part two of the lemma.
Philosophy: if this is too much arrow-ception you might try skipping these two pages and just reading the statement of the universal coefficient theorem and doing a couple exercises, then coming back to it. It's one of those theorems that you kind of have to prove for yourself instead of following someone else's proof because it's 8 million steps in a row, any individual one of which isn't too bad, but all of which together are confusing the first 7 to 12 times.