I found a few occurrences of the same proof about representability of elements of $H_2(M,\mathbb{Z})$ for $M$ a closed orientable smooth $4$-manifold. All of them stop at the very end claiming that the conclusion is clear. I found an analogue question on MO but even there it's not explained the last passage of the proof.
(from Kirby's Topology of $4$-Manifolds Theorem 1.1 page $20$) There is an isomorphism $$ H^2(M, \mathbb{Z})\cong [M,\mathbb{C}P^{\infty}]$$ so letting $\hat{\alpha}$ being the Poincaré dual of a chosen $\alpha \in H_2(M, \mathbb{Z})$, there is an homotopy class of maps $[f]\colon M \to \mathbb{C}P^{\infty}$ corresponding to $\hat{\alpha}$. By cellular approximation, we can homotopy (a representative of) $f$ in roder to obtain a map $f\colon M \to \mathbb{C}P^2$. Make $f$ smoothly transverse to $\mathbb{C}P^1\subset \mathbb{C}P^2$. Consider $f^{-1}(\mathbb{C}P^1)$, this will be an oriented surface representing $\alpha$.
I'm not able to prove this last statement, everything is pretty much clear. Looking around I didn't find any explanation for this, so I'm wondering if it is a trivial result. The only indication I found is in the linked question where one suggests to sue Pontrjagin-Thom construction for the group $SO(2)$. Needless to say I'm unable to make use of this hint.
Can someone give me an explanation for this last sentence?
The reason for this is the following fact:
Theorem: Let $N\to M$ be a submanifold and $[N]$ the homology class it defines in $M$. Then in $M$ the Thom class of the normal bundle is dual to $[N]$. (see e.g. this exercise in Milnor Stasheff)
Use this together with the map $[M,K(G,i)] \to H^i(M;G)$ which is precisely given by pulling back the unique $i$-cohomology class $k$.
Third thing we use is that normal bundles pull back and naturality of characteristic classes.
Now in your notation (let additionally $\tau_\alpha \in H^2M$ and $\tau_k\in H^2\mathbb CP^2$ be the Thom classes of the normal bundles of $f^{-1}(\mathbb CP^1) $ and $\mathbb CP^1$ resp.):
$$ \hat \alpha =f^*k = f^*(PD(\mathbb CP1)) = f^*(\tau _k) = \tau_\alpha. $$
Apply PD to this equality and obtain $$\alpha = PD(\hat \alpha)=PD(\tau_\alpha)=[f^{-1}(\mathbb CP^1)]$$.
Sorry for the very confusing notation which already kind of assumes the result.