A reflexive pair is a pair of morphisms $f, g: A \to B$ such that there exists a common section $s: B \to A$, i.e. $fs = gs = 1_B$.
What's the intuition behind this concept?
I tried looking at this in the category of sets to see if I recognised this as a well-known notion, but I don't see it. I see that $f$ and $g$ are split epis. I also constructed this basic example: $$ \begin{aligned}f : \{1,2,3\} &\to \{a, b\} \\ 1 &\mapsto a \\ 2 &\mapsto b \\ 3 &\mapsto b\end{aligned} $$ $$ \begin{aligned}g : \{1,2,3\} &\to \{a, b\} \\ 1 &\mapsto a \\ 2 &\mapsto a \\ 3 &\mapsto b\end{aligned} $$ This is a reflexive pair with section given by $a \mapsto 1$, $b \mapsto 3$. It seems any pair of parallel epis in $\mathrm{Set}$ is a reflexive pair, with common section given by selecting for every element in the codomain an element in the intersection of the preimage by one and the other morphism. However this didn't help in elucidating the concept.
To get something out of the way: not all pairs of parallel epis in $\mathsf{Set}$ have a common section. A simple example is where $E = \{a, b, c, d\}$ maps epimorphically to $\{0, 1\}$ in two ways $f, g$, where $f^{-1}(0) = \{a, b\}$ and $g^{-1}(0) = \{c, d\}$.
The intuition I have for reflexive forks as such is focused less on what "they are like" for certain concrete categories, and more on why we should care or what makes them so wonderful. Of course one big source for them is equivalence relations $E \hookrightarrow X \times X$, where by composing with the two product projections we get a pair of maps $E \underset{q}{\overset{p}{\rightrightarrows}} X$ with a common section $i: X \to E$, purely and simply because of the reflexivity property (that the diagonal map $\delta: X \to X \times X$ factors through the inclusion $E \hookrightarrow X \times X$), and this connection with the reflexive property is surely the reason for the name "reflexive pair". So you can and should think of reflexive forks as a useful generalization of equivalence relations.
As I'm sure you know, one of the primary things we do with an equivalence relation is pass to the quotient $q: X \to X/E$ formed by taking equivalence classes, and that's also one of the primary things we do with a reflexive fork $X \overset{i}{\to} E \underset{q}{\overset{p}{\rightrightarrows}} X$ as well: take the coequalizer of $p$ and $q$. For various reasons which I'll get into, coequalizers of reflexive pairs tend to behave a lot better than general coequalizers, and for many purposes, the nice features of coequalizers of equivalence relations are attributable just to reflexivity, without the extra overhead of symmetry and transitivity to take into account.
One quick observation is for a surprising number of algebraic situations, for example for the categories of groups, rings, and Heyting algebras among others, an internal relation $E \hookrightarrow X \times X$ is an equivalence relation if and only if it is reflexive. See the nLab article Mal'cev variety for discussion of this point.
Another quick observation is that a great number of coequalizers that arise in practice are in fact reflexive coequalizers. For example, you may know that to construct general colimits in a category, it suffices to be able to construct coproducts and coequalizers. But did you know that it suffices to construct coproducts and reflexive coequalizers? After all, when you sit down and examine how the coproducts-and-coequalizers construction works, say for constructing the colimit of a diagram $D: J \to C$, it's a coequalizer of type
$$\sum_{f: j \to k} Dj \rightrightarrows \sum_{j} Dj \to \mathrm{colim} D$$ where the parallel arrows have a common section (hint: consider the sum over $j$ as amounting to summing over identity arrows $f = 1_j$).
Somewhat relatedly, in a monadic situation (like the category of groups being monadic over the category of sets, with monad $M$), the canonical generators and relations presentation of an algebra $A$ with algebra structure map $\alpha: MA \to A$,
$$MMA \underset{M\alpha}{\overset{\mu A}{\rightrightarrows}} MA \overset{\alpha}{\to} A,$$ is a reflexive coequalizer, with common section given by $MuA: MA \to MMA$ (courtesy the unit $u: 1 \to M$ of the monad).
One relevant thing we can say in these situations is that a parallel pair of maps in a category looks like a fragment of a simplicial object where we just take into account the two face maps $C_1 \rightrightarrows C_0$ at the tail end, whereas a reflexive fork takes into account the degeneracy $C_0 \to C_1$ as well. Thus, the shape of a reflexive fork is (opposite to) the full subcategory of the simplex category whose objects are $0$ and $1$, with the degeneracy often playing an essential role.
But I still haven't said why reflexive coequalizers behave better than general coequalizers. One important clue in this respect is that the shape category (i.e., the full subcategory $\Delta_{0, 1}^{op}$ from the previous paragraph) has the property that the diagonal functor
$$\delta: \Delta_{0, 1}^{op} \to \Delta_{0, 1}^{op} \times \Delta_{0, 1}^{op}$$ is a final functor, meaning effectively that the colimit of any functor $F: \Delta_{0, 1}^{op} \times \Delta_{0, 1}^{op} \to C$ is isomorphic to the colimit of the functor $F\delta: \Delta_{0, 1}^{op} \to C$. (Categories with this property are called sifted categories.) This is not true of the shape category for a pair of parallel maps, and this accounts for a lot.
Why is this useful? Well, for example, we can exploit this fact to prove that in a category with products $C$, if $c \times -: C \to C$ preserves colimits say, then the product functor $- \times -: C \times C \to C$ preserves reflexive coequalizers. This fact is simply not true for general coequalizers!
This fact alone turns out to be a secret reason why so many functors arising in practice tend to preserve reflexive coequalizers but not so much general coequalizers, and that's generally why we care about reflexive coequalizers so much: their preservation properties. For example, the forgetful functor $\mathsf{Grp} \to \mathsf{Set}$ preserves reflexive coequalizers. I probably owe you an example that shows it doesn't preserve general coequalizers, but IIRC that's the case. It's also the underlying reason for why the crude monadicity theorem, which traffics in reflexive coequalizers, is far more useful in practice than the precise monadicity theorem. This is one of many examples.
This answer may be much longer than you bargained for, but these technical points are not always easily found in textbooks, so reflexive forks and their coequalizers are something like a secret weapon in the arsenal of category theorists.