Let me start with a remark : to check whether a given [EDITED for clarity after Andrew's comments]. Given $h:X\to Y$, the following are equivalent:
(1) $h$ is the coequalizer of some $W\rightrightarrows X$you can forget about the particular ,
(2) $W$, because you can then take h$is the coequalizer of$W=X\times_Y X\times_Y X\rightrightarrows X$with the two projections. In other words, being a coequalizer is equivalent to being an effective epimorphism (This works in any category with fiber products). Back to the questions. Question (b) asks whether if$h$is a coequalizer, then its restriction$h^{-1}(V)\to V$also is, for each open$V\subset Y$. Let me recall the example I gave to answer this question, which provides a counterexample where$h^{-1}(V)$is empty (and$V$isn't): take$Y=\mathrm{Spec}\,k[[t]]$($k$a field),$X=$the disjoint sum of all subschemes$\mathrm{Spec}\,(k[[t]]/(t^n))$($n\geq1$),$V=$generic point of$Y$. For question (a), assume each$h_i:X_i\to Y_i$is a coequalizer and let$s:X\to S$be a morphism such that$sf=sg$. Then for each$i$, the restriction of$s$to$X_i$descends uniquely to$t_i:Y_i\to S$. The question is whether$t_i$and$t_j$coincide on$Y_i\cap Y_j$. Composing them with (the restriction of)$f$(or$g$) gives the same result, hence: $\bullet$ gluing is automatic (and we get a positive answer) if we know that for each open$V\subset Y$, the restriction$h^{-1}(V)\to V$is an epimorphism of schemes; $\bullet$ but the above example shows that this is not true in general, and in fact we get a (nonseparated) counterexample to the question by taking two copies$X_i\to Y_i$($i=1,2$) of that example and putting$X=X_1\coprod X_2$,$Y=$gluing of$Y_1$and$Y_2$along the generic points: here the coequalizer of$X\times_Y X\rightrightarrows X$is$Y_1\coprod Y_2$. 1 Let me start with a remark: to check whether a given$h:X\to Y$is the coequalizer of some$W\rightrightarrows X$you can forget about the particular$W$, because you can then take$W=X\times_Y X$with the two projections. In other words, being a coequalizer is equivalent to being an effective epimorphism. Back to the questions. Question (b) asks whether if$h$is a coequalizer, then its restriction$h^{-1}(V)\to V$also is, for each open$V\subset Y$. Let me recall the example I gave to answer this question, which provides a counterexample where$h^{-1}(V)$is empty (and$V$isn't): take$Y=\mathrm{Spec}\,k[[t]]$($k$a field),$X=$the disjoint sum of all subschemes$\mathrm{Spec}\,(k[[t]]/(t^n))$($n\geq1$),$V=$generic point of$Y$. For question (a), assume each$h_i:X_i\to Y_i$is a coequalizer and let$s:X\to S$be a morphism such that$sf=sg$. Then for each$i$, the restriction of$s$to$X_i$descends uniquely to$t_i:Y_i\to S$. The question is whether$t_i$and$t_j$coincide on$Y_i\cap Y_j$. Composing them with (the restriction of)$f$(or$g$) gives the same result, hence: $\bullet$ gluing is automatic (and we get a positive answer) if we know that for each open$V\subset Y$, the restriction$h^{-1}(V)\to V$is an epimorphism of schemes; $\bullet$ but the above example shows that this is not true in general, and in fact we get a (nonseparated) counterexample to the question by taking two copies$X_i\to Y_i$($i=1,2$) of that example and putting$X=X_1\coprod X_2$,$Y=$gluing of$Y_1$and$Y_2$along the generic points: here the coequalizer of$X\times_Y X\rightrightarrows X$is$Y_1\coprod Y_2\$.