Composition of flat morphisms is flat

Question

I do not understand in the snippet below why the composition of flat morphisms is flat. The full text (though in Czech) is given here. I do not understand these step given in bold, beginning from $q\cdot u=0$ implies $q\cdot p_2\cdot g=0$, so $q \cdot p_2\cdot g=0$ so $q\cdot p_2=h\cdot p_3$ for some $h$. Then $$h\cdot v \cdot p_2=h\cdot p_3=q\cdot p_2$$ then $h\cdot v=q$.

Answer 1

I believe, after we have proven that $u$ is injective, we want to realise $C/A$ as an extension of flat modules $B/A$ and $C/B$. To this end, we first prove that $u$ is injective and then consider the quotient map $q\colon C/A\to C/B$. As $q\circ u= 0$ by construction, $0= q\circ u\circ p_1 = q\circ p_2\circ g$. Therefore $q\circ p_2$ vanishes on the image of $B$ in $C$ under the inclusion $g$ and hence factorises through $C/B$, so $q\circ p_2 = h\circ p_3$ for some $h\colon C/B\to C/B$.

Now, $h\circ v\circ p_2 = h\circ p_3 = q\circ p_2$. As $p_2$ is an epimorphism, it can be cancelled if it is the rightmost morphism, so $h\circ v = q$. This realises $C/A$ as an extension of flat modules $B/A\to C/A\to C/B$ with the inclusion map $u$ and the quotient map $q$.