The proof in Morgan's book seems to be incorrect. It is not true that every class $\alpha \in H_3(X;\mathbb Z/2^k\mathbb Z)$ is represented by a map from a smooth $\mathbb Z/2^k \mathbb Z$-oriented manifold into $X$.
Suppose $k>1$. Since being $\mathbb Z/2^k \mathbb Z$-oriented is the same as being oriented, $\alpha$ would need to be the mod $2^k$ reduction of an integral class. This is certainly not always the case.
For a specific example we can let $X=L_4^3 \times S^1$ where $L^3_4$ denotes the quotient of $S^3 \subset \mathbb C^2$ by the diagonal action of $\{\pm 1,\pm i\}\cong \mathbb Z/4\mathbb Z$. Then $H_3(X;\mathbb Z/4 \mathbb Z) \cong \mathbb Z/4 \mathbb Z \oplus \mathbb Z/4 \mathbb Z$ with the first summand generated by the fundamental class of $L^3_4$ and the second coming from $\mathrm{Tor}(H_2(X),\mathbb Z/4\mathbb Z)$. The second summand is not the reduction of an integral class and therefore can not be realized by a map from a $3$-manifold.
See Proposition 5.7.4 in Gompf and Stipsicz, 4-manifolds and Kirby calculus, for the original Hirzebruch-Hopf proof as well as a description of the Teichner-Vogt proof for arbitrary oriented $4$-manifolds linked to in the comments on the question.