$\require{AMScd}$ Define groups and homomorphismes as below. I am supposed to show, that given the left and the right rectangle are pushouts/amalgames, then the outer/big one is a pushout/amalgame aswell.
$$ \begin{CD} G @>f>> G' @>f'>> G''\\ @VgVV @VVg'V @VVg''V \\ H @>h>> H' @>h'>> H'' \end{CD} $$
Does that mean that I have to show that the following diagram is a pushout?
$$ \begin{CD} G @>f' \circ f>> G''\\ @VgVV @VVg''V \\ H @>h' \circ h>> H'' \end{CD} $$
If so, do I have to show that the different pushouts $H''$ of the right rectangle and the big rectangle coincide? I.e.
Given $H'$ and $H''$ are pushouts, they are defined as $H':= G'*_{G}~H$ and $H'':= G'' *_{G'}~H'$.
Given my previous thought process do I now have to show that (where $\newcommand{\ngenl}{\mathopen{\vartriangleleft}} \newcommand{\ngenr}{\mathopen{\vartriangleright}}\ngenl S \ngenr$ denotes the normal subgroup generated by $S$): $$ \begin{align*}G''*H/ \ngenl (f'\circ f)(x)g(x)^{-1} \forall x \in G\ngenr & =: G'' *_{G}~H \\ & \stackrel{!}{=}G'' *_{G'}~H'\\ &= G'' *_{G'}~(G' *_{G}~H)\\ &= \frac{G''*(G'*H/\ngenl f(x)g(x)^{-1} \forall x\in G\ngenr)}{\ngenl f'(\bar{x})g'(\bar{x})^{-1} \forall \bar{x}\in G'\ngenr} \end{align*} $$
And if so how do I continue ?
Unfortunately I do not understand at all what your formula with the many triangles at the end means. But it is probably simpler than what you are trying to do. You just check that the outer rectangle has the universal property of a pushout.
Let $T$ be any group. Using that both little squares are pushouts you can make the following computation. \begin{align*} Hom(H'',T) &= \{(a,b)\in Hom(G'',T)\times Hom(H',T)\mid a\circ f' = b\circ g'\} \\&= \{(a,c,d)\in Hom(G'',T)\times Hom(G',T) \times Hom(H,T)\mid d\circ f = c\circ g\wedge a\circ f' = (c,d)\circ g'\} \\&=\{(a,c,d)\in Hom(G'',T)\times Hom(G',T) \times Hom(H,T)\mid d\circ f = c\circ g\wedge a\circ f' = d\} \\&= \{(a,c)\in Hom(G'',T) \times Hom(H,T)\mid a\circ f'\circ f = c\circ g\} \end{align*}
I have inserted a few extra-steps in the middle. It is less complicated than it looks like. Just start with two arrows $c:H\to T$ and $a:G''\to T$ such that $c\circ g = a \circ f'\circ f$, and try to find an induced arrow $H''\to T$ using the universal properties of the two squares. Then you check that the arrow you get is the one and only arrow which fits, and you are done.