Abelianization of free product is the direct sum of abelianizations

Let $A$ be an abelian group. Then $$\begin{eqnarray*} \text{Hom}(G_1\ast G_2, A) & = & \text{Hom}(G_1,A)\times \text{Hom}(G_2,A) \\ & = & \text{Hom}(Ab(G_1),A)\times \text{Hom}(Ab(G_2),A) \\ & = & \text{Hom}(Ab(G_1)\oplus Ab(G_2),A) \end{eqnarray*}$$ and there you have your universal property.

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Explicitly, for any group $G$ I write $[g]_G$ for the class of $g\in G$ in $Ab(G)$. Then you get $\varphi: Ab(G_1\ast G_2)\to Ab(G_1)\oplus Ab(G_2)$ defined by $\varphi([g_1h_1g_2h_2\cdots g_rh_r]_{G_1\ast G_2}) = \left( [g_1\cdots g_r]_{G_1}, [h_1\cdots h_r]_{G_2}\right)$ with $g_i\in G_1$, $h_i\in G_2$.