In Hatcher page 192, the sequence is extracted
$B_n^* \leftarrow Z_n^* \leftarrow H^n(C;G) \leftarrow B_{n-1}^* \leftarrow Z_{n-1}^*$ from the diagram
$Z_n^* = \ker \delta_n$, $B_n^* = Img \delta_{n-1}$ are the dual spaces of the spaces without $*$
Let explain as I myself found it terse. This dual map is obtained from the split exactness of the row. This is split because $B_n$ is free. So the "pulling" (not pullback) described is from the split exactness. I am not quite sure what he meant by undoing here, but in any case that is the construction.
Basically my question is, how do we get the $H^n(C;G)$ in the middle of the sequence $B_n^* \leftarrow Z_n^* \leftarrow H^n(C;G) \leftarrow B_{n-1}^* \leftarrow Z_{n-1}^*$?
As far as I can tell, this is just part of the long exact sequence of homology associated to the diagram (ii). In (ii) the homology of the vertical column at $C_n^*$ is $H^n(C;G)$ (I presume Hatcher defined $H^n(C;G)$ thusly). The homology of the left column is just the $Z_n^*$, since the differentials are zero. Similarly for the right column.