Suppose you have two morphisms $f\colon A\to B$ and $g\colon A\to C$. Then the fiber coproduct of $B$ and $C$ over $A$ exists and is the disjoint union $B\coprod C$ where we identify $fx$ and $gx$ for all $x\in A$.
I'm a little confused on how this satisfies the universal property. Suppose you had another set $D$ and morphisms $h\colon B\to D$ and $k\colon D$ such that $hf=kg$, so that the diagram commutes. How would you define the unique morphism $u$ from $B\coprod C\to D$?
Is it something along the lines of $u(b)=h(b)$ if $b\notin f(A)$ and $u(c)=k(c)$ if $c\notin g(A)$, and $u(f(x))=u(g(x))=k(f(x))=h(g(x))$ if $f(x)=g(x)$ is one of the points we smashed together in $B\coprod C$? And this definition makes sense since $hf=kg$.
That is exactly it. Can you see why the map is unique?
(Perhaps it is better to call this just the "pushout" of $B \leftarrow A \rightarrow C$, or more informally the "glueing of $B$ and $C$ along $A$". The name "fibre coproduct" seems a bit weird, because the construction has nothing to do with fibres!)