For questions like this it is convenient to take the functor-of-points view of schemes. Fix a base scheme $S$, and let $h : Z \to X$ be the equaliser of $f, g : X \to Y$ in $\textbf{Sch}_S$. This exists since $\textbf{Sch}_S$ has fibre products and a terminal object. (It is indeed the pullback of the diagonal $\Delta_Y : Y \to Y \times_S Y$ along $\langle f, g \rangle : X \to Y \times_S Y$, but there are other descriptions.) By the universal property of equalisers, there is a bijection $\DeclareMathOperator{\Spec}{Spec}$ $\textbf{Sch}_S(T, Z) \cong \{ \bar{x} \in \textbf{Sch}_S(T, X) : f \circ \bar{x} = g \circ \bar{x} \}$ and this is natural in $T$. Thus, the $T$-valued "points" of $Z$ are exactly what you expect them to be. In particular, if $S = T = \Spec k$ is an algebraically closed field, and $X$ is a scheme locally of finite type over $\Spec k$, then the $k$-valued points of $Z$ are exactly the closed points of $X$ on which $f$ and $g$ agree; indeed, under these hypotheses, the residue field of a closed point of $X$ is guaranteed to be canonically isomorphic to $k$, so there is a natural bijection between $\textbf{Sch}_k(\Spec k, X)$ and closed points of $X$. (I'm not sure whether all closed points of $Z$ are $k$-valued. This at least seems plausible, since it is true in the case where $X$ and $Y$ are affine.)
In general, however, we must work a little harder. By general abstract nonsense, there is a canonical map from the "underlying" points of $Z$ to the set $\{ x \in X : f (x) = g (x) \}$. Is it surjective? Let $x$ be a point in $X$ such that $y = f (x) = g (x)$. Let $f^\sharp_x, g^\sharp_x : \mathscr{O}_{Y, y} \to \mathscr{O}_{X, x}$ be the induced homomorphisms on local rings, and form the coequaliser $\mathscr{O}_{Y,y} \rightrightarrows \mathscr{O}_{X,x} \rightarrow B$ in the category of commutative rings. The homomorphism $\mathscr{O}_{X,x} \to B$ is guaranteed to be surjective, so assuming $B$ is not the trivial ring, $B$ is a local ring and $\mathscr{O}_{X,x} \to B$ is a local homomorphism. The residue field of $B$ is canonically isomorphic to $\kappa (x)$, so we get an $S$-morphism $\bar{x} : \Spec \kappa (x) \to X$ such that $f \circ \bar{x} = g \circ \bar{x}$, and it lifts to an $S$-morphism $\bar{z} : \Spec \kappa (x) \to Z$; if $z$ is the underlying point of $\bar{z}$, then we have $h(z) = x$, as required.
However, if $B$ is the trivial ring, all hell breaks loose, and in that case $x$ is not in the set-theoretic image of $h : Z \to X$. For, if $x = h(z)$, then there would be a diagram of local rings $\mathscr{O}_{Y,y} \rightrightarrows \mathscr{O}_{X,x} \rightarrow \mathscr{O}_{Z,z}$ and $h^\sharp_z : \mathscr{O}_{X,x} \to \mathscr{O}_{Z,z}$ must factor through $\mathscr{O}_{X,x} \to B$ by the universal property of $B$ as a coequaliser. But then $\mathscr{O}_{Z,z} = 0$ – a contradiction.
This situation is not impossible either. For example, take $S = \mathbb{F}_p$, $X = Y = \Spec \overline{\mathbb{F}_p}$, $f = \textrm{id}$, and $g$ the Frobenius $p$-power map $t \mapsto t^p$. Then the equaliser of $f, g : X \to Y$ is the empty scheme, because the coequaliser of the corresponding homomorphism of rings is the trivial ring.