I can't speak for Matt Emerton specifically, but my understanding is that it is conventional to describe maps in terms of $k$ points in such a way that there is a clear extension of the definition to $S$ points. This is perhaps less rigorous, but if you know how to fill in the details, it removes clutter and leaves the essential geometric idea intact.

A partial justification for this practice is that a map between varieties over an algebraically closed field is determined by its value on $\overline k$ points, so in "nice" cases this description is not too lossy.

In your specific case, note that the $k$ point $x$ yields a constant $S$ point $S \to {\rm Spec} k \to \mathbb P^n$. There is a map that takes a pair of disjoint points of $\mathbb P^n$ to a rank $2$ linear subspace of $\mathbb P^n$. One way of describing it on $S$ points is as follows. Let $\mathcal O_S^{\oplus n+1} \to L_1$ and $\mathcal O_S^{\oplus n + 1} \to L_2$ be the surjections corresponding to two $S$ valued points $x_1, x_2$. Then dualizing, we obtain an injection $L_1^{\vee} \oplus L_2^{\vee} \to \mathcal O_S^{\oplus n+1}$ (this follows from Nakayama and the fact that $x_1, x_2$ are disjoint). Then take the subscheme of $\mathbb P^n_S$ defined by the vanishing of the homogeneous ideal of ${\rm Sym}(\mathcal O_S^{\oplus n+1})$ generated by $L_1^{\vee} \oplus L_2^{\vee}$. This yields the map $(\mathbb P^n-x)(S) \to {\rm Hilb}(\mathbb P^n)(S)$. Then pullback the subscheme along the flat map $Y \to \mathbb P^n$ to get the first map in Emerton's answer.

As you can see, this description more verbose. In my opinion, it obscures the key point: given two disjoint lines you can take their span and get a plane!

(One final nitpick: I don't think your dichotomy "old school" v.s. "modern" is very accurate. Classical algebraic geometers were comfortable with defining maps without always writing down explicit equations.)

I can't speak for Matt Emerton specifically, but my understanding is that it is conventional to describe maps in terms of $k$ points in such a way that there is a clear extension of the definition to $S$ points. This is perhaps less rigorous, but if you know how to fill in the details, it removes clutter and leaves the essential geometric idea intact.

A partial justification for this practice is that a map between varieties over an algebraically closed field is determined by its value on $\overline k$ points, so in "nice" cases this description is not too lossy.

In your specific case, note that the $k$ point $x$ yields a constant $S$ point $S \to {\rm Spec} k \to \mathbb P^n$. There is a map that takes a pair of disjoint points of $\mathbb P^n$ to a line in $\mathbb P^n$. One way of describing it on $S$ points is as follows. Let $\mathcal O_S^{\oplus n+1} \to L_1$ and $\mathcal O_S^{\oplus n + 1} \to L_2$ be the surjections corresponding to two $S$ valued points $x_1, x_2$. Then dualizing, we obtain an injection $L_1^{\vee} \oplus L_2^{\vee} \to \mathcal O_S^{\oplus n+1}$ (this follows from Nakayama and the fact that $x_1, x_2$ are disjoint). Then take the subscheme of $\mathbb P^n_S$ defined by the vanishing of the homogeneous ideal of ${\rm Sym}(\mathcal O_S^{\oplus n+1})$ generated by $L_1^{\vee} \oplus L_2^{\vee}$. This yields the map $(\mathbb P^n-x)(S) \to {\rm Hilb}(\mathbb P^n)(S)$. Then pullback the subscheme along the flat map $Y \to \mathbb P^n$ to get the first map in Emerton's answer.

As you can see, this description more verbose. In my opinion, it obscures the key point: given two disjoint lines you can take their span and get a plane!

(One final nitpick: I don't think your dichotomy "old school" v.s. "modern" is very accurate. Classical algebraic geometers were comfortable with defining maps without always writing down explicit equations.)