Proof:

To establish uniqueness, suppose that $G^{\prime}$ is another candidate for the coproduct with the associated homomorphisms $j_1:G_1\longrightarrow G^{\prime}$ and $j_2:G_2\longrightarrow G^{\prime}$ satisfying the universal property. Taking $f_1 = j_1$ and $f_2 = j_2$ in the definition, there exists a homomorphism $\phi:G\longrightarrow G^{\prime}$ such that

$$\phi\circ i_1 = j_1,\quad \phi\circ i_2 = j_2.$$

But since $G^{\prime}$ is also a coproduct we obtain reciprocally a group homomorphism $\psi:G^{\prime} \longrightarrow G$ such that

$$\psi\circ j_1 = i_1,\quad \psi\circ j_2 = i_2.$$

Combining the two we get $(\psi\circ\phi)\circ i_1 = i_1$ and $(\psi\circ\phi)\circ i_2 = i_2.$ We see that the identity map id$_G$ as well as $\psi\circ\phi$ satisfy the universal property with $H = G$, $f_1 = i_1$ and $f_2 = i_2$. The uniqueness clause in the definition of the coproduct gives $\psi\circ\phi =$   id$_{G}$ Interchanging the roles of $G$ and $G^{\prime}$ we get $\phi\circ\psi =$   id$_{G^{\prime}}.$ We leave it to the student to show that the maps $i_1$ and $i_2$ are injective.