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?
The universal property implies that the map must be a one-to-one set-theoretic map.
To see this, let $a,b\in S$ be such that $g(a)=g(b)$. Let $G$ be a nontrivial group (e.g., $G=C_2$, the cyclic group of order $2$) and let $g\in G$ be a nontrivial element. Let $f\colon S\to G$ be defined by
$$f(s) = \left\{\begin{array}{ll}
1 & \text{if }s\neq a,\\
g &\text{if }s=a.
\end{array}\right.$$
By the universal property, there exists a group homomorphism $\varphi\colon F\to G$ such that $f=\varphi\circ g$. In particular, $f(b) = \varphi(g(b)) = \varphi(g(a)) = f(a) = g$, hence $b=a$ (since the only element of $S$ that is mapped to $g$ by $f$ is $a$).
Therefore, $g$ is one-to-one.
Once you know it is one-to-one, you may replace $S$ with $g(S)$ and consider it to be the inclusion, since the universal property also gives:
Theorem. Let $S$ and $T$ be sets, and let $f\colon S\to T$ be a bijection. If $(g,F_S)$ and $(h,F_T)$ are free groups on $S$ and on $T$, then $f$ induces a unique isomorphism $\Phi\colon F_S\to F_T$ such that $\Phi\circ g = h$ and $\Phi(g(s)) = h(f(s))$ for all $s\in S$.
Proof. Use the universal property of $(g,F_S)$ with $h\circ f$ to obtain $\Phi$. Then use the universal property of $(h,F_T)$ with $g\circ f^{-1}$ to obtain a map $\Psi$. Finally, use the uniqueness clause of the definition to prove that $\Phi\circ\Psi$ and $\Psi\circ\Phi$ are the corresponding identity morphisms. $\Box$
So we can replace a free group on $S$ $(g,F_S)$ with the free group $(\iota,F_{g(S)})$ which is free on $g(S)$, and which is canonically isomorphic to $(g,F_S)$.
Best Answer
First of all, this universal property result answers a very natural question, namely the question of how the homomorphisms $G/H \to L$ look like. In fact, by the universal property, there is a one-to-one correspondence between the homomorphisms $G/H \to L$ and the homomorphisms $G \to L$ having $H$ in its kernel and this correspondence is given by composing with $\pi$.
Secondly, an object having a universal property is unique with this property up to isomorphy. In your case this means that if $\pi': G \to L'$ is a homomorphism with $H$ in its kernel and the property that for every $\varphi: G \to L$ which has $H$ in its kernel we have a factorization $\varphi = \varphi' \circ \pi'$ for a unique $\varphi': L' \to L$, then there exists an isomorphism $\psi : G/H \cong L'$ such that $\pi' = \psi \circ \pi$.
As an example of how this can be used, let us consider quotients of quotients. If $N$ is another normal subgroup of $G$ contained in $H$, then $H/N$ is normal in $G/N$ and we want to consider $(G/N)/(H/N)$ which comes equipped with a canonical map $\pi_H: G/N \to (G/N)/(H/N)$. Let $\varphi: G \to L$ be a map with $H$ in its kernel, then it also contains $N$ in its kernel and we get a factorization $\varphi = \varphi' \circ \pi_N$ for a unique $\varphi' : G/N \to L$ where $\pi_N : G \to G/N$ is the canonical projection. Now, the homomorphism $\varphi'$ has $H/N$ in its kernel (this follows from the above factorization), hence we get a unique $\varphi'' : (G/N)/(H/N) \to L$ with $\varphi'' \circ \pi_H = \varphi'$, thus $\varphi''$ is unique with $\varphi'' \circ (\pi_H \circ \pi_N) = \varphi$, so $(G/N)/(H/N)$ has the same universal property as $G/H$ (with respect to the morphism $\pi_H \circ \pi_N$), hence $G/H \cong (G/N)/(H/N)$. If you would show this without using the universal property you would have to define the isomorphism directly and show that it is well-defined (independent of the coset representative), a homomorphism etc. This is not that difficult. But in my opinion, the proof using the universal property also tells you why these groups are isomorphic the reason being that they satisfy the same universal property.