Let $Ho(Ch_Z)$ be the localization of the category of nonnegatively graded complexes of abelian groups, $Ch_Z$, wrt the quasi-isomorphisms.
If two objects are isomorphic, in this localization, then are they quasi isomoprhic? I do not think this is true. But why in page 43 of DS's notes, we have two objects in $A,B \in Ho(Ch_Z)$ for which the author concludes $$H_iA \cong H_iB$$
I know this may be the case when $A,B$ are special objects. The problem is I cannot really tell if this is case in the notes.
Strictly speaking: no, as commented before, the zigzags make problems (assuming you already tamed them into defining sets of morphisms, which you can via classical construction and embedding your category in the category of bounded chain complexes), for example consider the following 2 complexes of abelian groups (i.e. $\mathbb{Z}$-modules): $$ 0\to 0\to \mathbb{Z}/2\mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{2} \mathbb{Z} \to 0 $$ and $$0\to \mathbb{Z} \xrightarrow{2} \mathbb{Z} \to 0\to \mathbb{Z}/2\mathbb{Z}\to 0 $$
then ther just can't be a quasi-imorphism going either way between those complexes (there are no nontrivial morphisms from $\mathbb{Z}/2\mathbb{Z}$ to $mathbb{Z}$). However, both are quasi-isomorphic to $$0 \to 0 \to \mathbb{Z}/2\mathbb{Z} \to 0 \to \mathbb{Z}/2\mathbb{Z}\to 0$$.
Regarding your question in the proof: as mentioned before in the comments, you none the less get a zigzag of quasiisomorphisms (in worsecase to a projective resoltion or a cofibrant fibrant replacement, depending on the setting you are working in), which then all induce isomorphisms of $H^*$, but those are now actual isomorphisms in the category of groups, hence you can actually invert them, and make the zigzag into a single iso.
Also, be aware that often people (especially when using model categories) restrict themselves to certain subcategories, for example $\mathcal{Ho}(\mathcal{C})$ usually is the subcategory of fibrant-cofibrant objects modulo homotopy.