Hatcher's proof of Quillen plus construction 
This is from Hatcher's Algebraic Topology. This Proposition says we can kill the fundamental group of an arbitrary connected CW complex with $H_1=0$, without affecting the homology groups. The proof was clear to understand, until the underlined statement. I see that $H_2(X')\cong \pi_2(X')$ by the Hurewicz theorem , but how can we represent a basis of $H_2(X',X)$ by maps $\psi_\alpha :S^2 \to X'$?
 A: You have a typo in your very last formula, it should be $\psi_\alpha : S^2 \to X'$ (not $\psi_\alpha : S^2 \to X$).
The Hurewicz homomorphism $\mathcal H : \pi_2(X') \approx H_2(X')$ (which is an isomorphism in this case where $X'$ is simply connected) is defined in the statement of the Hurewicz theorem. The definition is this: given an element $[\psi] \in \pi_2(X')$ represented by a continuous map $\psi : S^2 \to X'$, the homology class $\mathcal H[\psi] \in H_2(X')$ is defined to be the image of the standard generator of $H_2(S^2)$ under the induced homomorphism $\psi_* : H_2(S^2) \to H_2(X')$. Since $\mathcal H$ is an isomorphism in this case, it follows that every element of $H_2(X')$ can be represented in the form $\mathcal H[\psi]$ for some appropriate choice of $\psi$. 
So, pick $\{\alpha_i\}$ to be a basis for $H_2(X',X)$, let $\{\alpha'_i\} \subset H_2(X')$ be the image of that basis under the splitting, and so you may choose $\psi_{\alpha_i} : S^2 \to X'$ which represents $\alpha_i$ under the Hurewicz homomorphism as just described. 
