Let $s, t:B\rightarrow C$ be the cokernel pair.
$C$ is the coproduct $B + B$, but with the image of $f$ identified (glued together), in other words:
$$C = \mathrm{im} f + (B - \mathrm{im} f) + (B - \mathrm{im} f)$$
The two morphism $s, t$ are simply the left and right coprojections into $C$.
To see why this is, observe that the cokernel pair is defined as the pushout of the function along itself:
$$ \array{ &&&& A &&&& \\\ & && f \swarrow & & \searrow f && \\\ && B &&&& B \\\ & && s\searrow & & \swarrow t && \\\ &&&& C &&&& } $$
We know that in **Set** , $C$ is defined as the set $B + B$ up to the equivalence $b_1 \cong b_2$ iff there exists $a \in A$ such that $f(a) = b_1$ and $f(a) = b_2$; in other words, iff $b_1 = b_2 \in \mathrm{im} f$.
This means that $C$ has one copy of $\mathrm{im} f$ and two copies of every other element.