These two group theory statements are "the same"?
By the Homomorphism Theorem, any homomorphism $f\colon G\to K$ factors through $G/\mathrm{ker}f$, meaning that there is a map $\hat{f}\colon G/\mathrm{ker}f \to K$ such that $ f = \hat{f}\pi$. The map is indeed $\hat{f}(g\,\mathrm{ker}f) = f(g)$. This applies to the specific case given in Statement 2.
Edit: In fact, the full Statement 1 is not equivalent to Statement 2, in the sense that if you replace $G'$ with an arbitrary subgroup $M$ of $G$ in both statements, then Statement 1 characterizes $G'$, but Statement 2 does not. That is, if you have
Statement 1': If $H\triangleleft G$ and $G/H$ is abelian, then $M\subseteq H$; and if $M\subseteq H$, then $H\triangleleft G$ and $G/H$ is abelian.
Statement 2': If $\varphi\colon G\to A$ is any homomorphism of $G$ into an abelian group $A$, then $\varphi$ factors through $M$; that is, $M\subseteq \ker\varphi$ and there is a homomorphism $\hat{\varphi}\colon G/M\to A$ such that $\varphi(g) = \hat{\varphi}\circ\pi(g)$.
The only subgroup $M$ of $G$ that satisfies Statement 1' is $M=G'$. However, any subgroup of $G'$ that is normal in $G$ will satisfy Statement 2'.
In fact, Statement 2 is equivalent to the first clause of Statement 1, namely that if $H\triangleleft G$ and $G/H$ is abelian, then $G'\subseteq H$, plus the implicit assertion that $G'$ itself is normal in $G$.
Assuming the first clause of Statement 1 plus the fact that $G'\triangleleft G$, if $\varphi\colon G\to A$ is a homomorphism, then by the Homomorphism Theorem, letting $H=\ker\varphi$, then $G/H$ is (isomorphic to) a subgroup of $A$, hence abelian, so we must have $G'\subseteq H = \mathrm{ker}\varphi$; this is Statement 2 (with the final clause of 2 given by the homomorphism theorem as above).
Assuming Statement 2, (which implicitly asserts that $G'$ is normal) suppose that $H$ is a normal subgroup of $G$ such that $G/H$ is abelian. Then considering $\pi\colon G\to G/H$ and applying 2, you conclude that $G'\subseteq \mathrm{ker}\pi = H$. And normality of $G'$ follows from the statement of 2, which requires it.
That is, they aren't quite equivalent, because Statement 1 has another clause, namely the "Conversely..." clause, which is not a consequence of assuming Statement 2. But the first part of Statement 1 (plus "$G'\triangleleft G$") is equivalent to Statement 2.
Added earlier: To see that the two are not quite equivalent as stated, let me give you an example of a subgroup $M$ of $G$ that satisfies Statement 2' but not Statement 1': consider the case of $G=S_4$; then $G' = A_4$. Now let $M = \{ 1, (12)(34), (13)(24), (14)(23)\}$. Then $M\triangleleft G$, and the statement in 2 holds for $M$: given any homomorphism $f\colon G\to A$ with $A$ abelian, the map $f$ factors through $G/M$ and there exists a homomorphism $\hat{f}\colon G/M\to A$ such that $f=\hat{f}\pi$. However, $M$ is not the commutator subgroup of $G$. What is missing in Statement 2 for it to be a true equivalent of Statement 1 is some statement that corresponds to the assertion that $G/G'$ is itself abelian, which is what follows from the "Conversely..." clause in Statement 1. One way to do it is to simply state that $G/G'$ is itself abelian. Another is to consider the intersection of all kernels of all homomorphisms into abelian groups, and say that $G'$ must be equal to that intersection.
Your idea for (2) exactly works. Any two representatives $a$, $b$ with $aG' = bG'$ will be related by an element $g' \in G'$ with $ag' = b$. Then, because $G'$ is a subset of the kernel of $\varphi$, $$\varphi(b) = \varphi(ag') = \varphi(a)\varphi(g') = \varphi(a).$$
This really is a restatement of the first isomorphism theorem for groups to a "first homomorphism theorem for groups", so to speak. The isomorphism theorem says that map $\theta: G \to A$ with kernel $K$ will factor through the quotient map $G \to G/K$ --- but really for any subgroup $N$ of $K$ with $N$ normal in $G$ will give a map $G \to G / N$ by the same recipe described above.
In your specific case, when the target is abelian, the kernel necessarily contains the commutator subgroup $G'$ of $G$, and so you use the "first homomorphism theorem" to get your map $\hat\varphi: G / G' \to A$.
That the first isomorphism theorem can be weakened in this way means that, whereas the first isomorphism theorem alone relates surjective maps off $G$ to normal subgroups of $G$, the weakening says that the inclusions $N_1 \subseteq N_2$ of normal subgroups of $G$ is also reflected on the level of homomorphisms.
Hope this helps!