Here, I found the proof of presentation of the free product of groups. I wanted to show the same thing for amalgamated free product of 2 groups i.e.
Show that that if $H$ is generated by $S$, and $ \phi_1 : H \rightarrow G_1 , \phi_2 : H \rightarrow G_2$ are morphisms of groups then the amalgamated product of $G_1$ and $G_2$ above $H$ is
$$ G_1 *_{H} G_2 = \langle S_1,S_2\; ; \; R_1,R_2,\phi_1(s) ( \phi_{2}(s) )^{-1},s \in S \rangle $$
where $G_1 = \langle S_1 \; ; \; R_1 \rangle$ and $G_2 = \langle S_2 \; ; \; R_2 \rangle$.