In Emily Riehl, "Category theory in Context", Section 4.6: "Existence of adjoint functors", the free group on a set $S$ is constructed by specializing the proof of the general adjoint functor theorem to the situation at hand. There are group-theoretic details of the proof which I don't follow. I shall run through the proof and point out details that I'd like explanations for.
Construction
Start with the forgetful functor $U: \textbf{Group} \to \textbf{Set}$, and a se $S$. Let $\Phi'$ be the set of representatives for isomorphism classes of groups which have a generating set of size at most $|S|$. Let $\Phi$ be the set of functions $f_G: S \to UG$ for each $G \in \Phi'$, where the image of $f_G$ is a generating set for $G$ (This is possible since every $G \in \Phi'$ has a generating set of size at most $|S|$).
We form a product:
$$
S \xrightarrow{\hat \eta} U \left( \prod_{S \to UG \in \Phi} G \right)
$$
(Question 1: What is the precise definition of the map $\hat \eta$?)
We define $\Gamma$ to be the subgroup $\Gamma \subset \prod_{S \to UG \in \phi} G$, where $\Gamma$ generated by the image of $\hat \eta$. We claim that $\Gamma$ is free since the map $\eta: S \to U\Gamma$ is initial in the category of maps $S \downarrow U$. (Question 2: Why does the map being initial mean that $\Gamma$ is free?)
Recall that for the map $\eta: S \to U \Gamma$ to be initial in $S \downarrow U$, any other map $\psi: S \to UH$ must factor through $\eta$. Let $G \subseteq H$ be a subgroup of $H$ that is generated by $S$. Thus we must have a map $\phi: S \to UG \in \Phi$. We claim that we can factor $\psi$ through $\phi$ by projection:
$$
S \xrightarrow{\phi} G \hookrightarrow H ~~\text{factorizes}~~ S \xrightarrow{\psi} H \\
S \xrightarrow{\eta} \prod_{S \to UG \in \Phi} G \xrightarrow{\pi_G} G \hookrightarrow H ~~\text{factorizes}~~ S \xrightarrow{\psi} H \\
$$
(Question 3: $\phi$ has image $G$, not $H$, while $\psi$ has image $H$. So why is $\phi$ a factorization of $\psi$?)
Example 4.6.6 screenshot
Best Answer
For each $G$, you have a map $f_G\colon S\to U(G)$, the underlying set of $G$, given by sending $S$ ot a generating set of $G$ (you "really" have one such map for each $G$ and each generating set and each set function from $S$ to the generating set).
Because you have a map from $S$ to each $UG$, the universal property of the product (in $\mathsf{Set}$) gives you a unique map from $S$ to $\prod_{G\in\Phi}UG$. But the product of the underlying sets is the same as the underlying set of the product, so this is the same as a map from $S$ to $U(\prod_{G\in\Phi}G)$. This maps sends $s$ in $S$ to the element whose $G$-component is $f_G(s)$. This is the map $\hat{\eta}$.
The universal property of the free group on $S$, $\Gamma$, is that for every group $G$, if you have a map $S\to U(G)$, this gives a unique map from $\Gamma$ to $G$ that suitably "extends" the map $S\to UG$. If you write down what the universal property of the initial object of $S\downarrow U$, you will see that it is exactly the universal property of the free group on $S$. The fact that $\eta$ is initial in the suitable category yields precisely that $\Gamma$ has the universal property of a free group, hence is free.
You need to compose with the inclusion $G\hookrightarrow H$. This is usually omitted, since it is just the fact that you are mapping to a subgroup of $H$; we don't usually think of, for example, the map $C_2\to S_3$ mapping onto $\{e,(12)\}$ as consisting of a map into the group $\{e,(12)\}$ followed by the inclusion of that group into $S_3$: we just think of the map being a map into $S_3$. The same is being done here: you think of it as mapping to $G$ to get the factorization, and then formally compose it with the inclusion $G\to H$ to "view it" as a map into $H$.