Pushout of categories in Set I found this answer Understanding an Example of a Pushout in  very helpful regarding disjoint unions, but wanted to know how to understand a pushout of categories in Set, say from : A→B and : A→C, where f and g are functors with shared domain A. There is an equivalence relation on objects, but how to deal with morphisms?

As you say, the pushout $P$ has objects $\mathrm{ob}(P)=\mathrm{ob} B\sqcup \mathrm{ob} C/\sim$, where $\sim$ is the equivalence relation generated by the relation containing $(b,c)$ whenever $b$ and $c$ are the image of the same object in $A$. Now to construct morphisms, if $p_1,p_2\in \mathrm{ob}(P)$, then we start with $$P'(p_1,p_2)=\bigsqcup_{p_1=[b_1],p_2=[b_2]} B(b_1,b_2)\sqcup \bigsqcup_{p_1=[c_1],p_2=[c_2]} C(c_1,c_2).$$
Now, $P'''$ is only a graph, not a category. We let $P''$ be the category freely  generated by $P'''$, so that a morphism $p_1\to p_2$ is a path $p_1\to p_2$ in $P'''$. 
Next, we construct $P'$ by imposing relations on $P''$ asserting that the mappings $B\to P',C\to P'$ are functors. Namely, we require that the path $(f,g)$ is identified with the path $(g\circ f)$ when the latter composition is defined in $B$ or in $C$ and that the paths $(\mathrm{id}_b)$ are identified with $\mathrm{id}_p$ for every $p\in \mathrm{ob} P$ and $b$ such that $p=[b]$. Constructed in this way, $P'$ is the pushout of $B$ and $C$ over $\mathrm{ob} A$. 
Finally, we construct $P$ itself by imposing further relations asserting that the two functors $A\rightrightarrows P$ must be equal. Now a functor $P\to Q$ is naturally identified with a graph morphism $P'''\to U(Q)$, where $U$ denotes the underlying graph, which respects the compositions and identities of $B$ and $C$ and coequalizes the two maps from $A$; that is, $P\to Q$ is uniquely identified with a pair of functions from $B$ and $C$ which are act equally on $A$. Thus $P$ has the desired universal property.
