We say that a homomorphism $f\colon G\to K$ "factors through" another homomorphism if one can write $f$ as a composition using that homomorphism.
For example, if $f\colon G\to K$ "factors through" $\pi\colon G\to H$, that means that there exists a homomorphism $u\colon H\to K$ such that $f = u\circ\pi$. The reason we call it "factors through" is that if you write composition of functions by simple juxtaposition, you get $f=u\pi$, which suggests that $\pi$ "divides" $f$, or that you can "factor" $f$ into a "product" in which one of the factors is $\pi$.
What the question is asking you to do is show that if $\phi\colon G\to A$ is a homomorphism from $G$ into an abelian group, then the homomorphism $\phi'\colon G/G' \to A/A'$ that is induced by $\phi$ (which is given by the formula $\phi'(gG')=\phi(g)A'$) satisfies $\phi(x) = \phi'(\pi(x))$ for all $x\in G$ (that is, that the function $\phi$ is the same as the function $\phi'\circ\pi$).
(I am assuming you have already shown that if $f\colon G\to K$ is a group homomorphism, then $f'\colon G/G'\to K/K'$ given by $f'(gG') = f(g)K'$ is a well-defined group homomorphism; if you haven't, then you need to do it!)
You got started correctly: technically, $\phi'\circ\pi$ cannot equal $\phi$, because the codomain of $\phi$ is $A$, while the codomain of $\phi'\circ\pi$ is $A/A'$. So your first step, showing that $A'$ is trivial, was great. That means that $A/A'$ is "really" (canonically) the same thing as $A$, so that you can consider $\phi'$ as being a map $\phi'\colon G/G'\to A$; thus, $\phi'$ "can be thought of" as given by $\phi'(gG') = \phi(g)$. So then you just need to verify the two functions, $\phi$ and $\phi'\circ\pi$, are equal.
The fact that every element of $G'$ maps to the trivial element of $A$ is important to show that $\phi'$ is well-defined, but if you already know that it is well-defined, then you don't need it.
Remember that the larger $N\lhd G$ is $G$ is, the smaller $G/N$ will be. Now suppose $N\lhd G$ and $G/N$ is abelian. This means that $abN=baN$ for any $a,b\in G$. This translates in $a^{-1}b^{-1}abN=N$. This means $N$ contains all commutators, and hence contains $[G,G]$. Thus, whenever $N$ is normal and the quotient $G/N$ is abelian $[G,G]\leq N$. Thus $G/[G,G]$ is as big as $G/N$ gets whenever $N$ is normal and $G/N$ is abelian.
Best Answer
If $f : G \to H$ is a homomorphism to an abelian group, then $f(ab) = f(a) f(b) = f(b) f(a) = f(ba)$, hence $[a, b] \in \ker f$, hence $[G, G] \subseteq \ker f$. Is the rest clear from here?