Proving that the free group on two generators is the coproduct $\mathbb{Z}*\mathbb{Z}$ in $\textbf{Grp}$

You're working too hard. I'll assume you're convinced that $\mathbb{Z}$ is the free group on one generator, which means you're convinced that

$$\text{Hom}(\mathbb{Z}, G) \cong G$$

(where the RHS should really be the underlying set of $G$ but I'm omitting that I'm applying the underlying set functor). One way to state the universal property of the coproduct is that

$$\text{Hom}(X \sqcup Y, Z) \cong \text{Hom}(X, Z) \times \text{Hom}(Y, Z)$$

from which it follows that

$$\text{Hom}(\mathbb{Z} \sqcup \mathbb{Z}, G) \cong G \times G$$

and this is also the universal property of the free group on two generators, so you're done by the Yoneda lemma. The argument is exactly the same for the free group on $n$ generators, and in fact on a set's worth of generators.

More abstractly, the forgetful functor $\text{Grp} \to \text{Set}$ has a left adjoint, the free group functor. As a left adjoint, it preserves colimits, and in particular coproducts. But every set $X$ is the coproduct of $X$ copies of the $1$-element set, so it follows that the free group $F_X$ is the coproduct of $X$ copies of $\mathbb{Z}$.