@Reuben asked:

> how do we make this definition of limit yield the notion of a product (which I understand is a kind of limit)

Let's see what happens when we set \\(\mathcal{D} = \textrm{2}\\) (the category with two discrete objects).

A \\(\textrm{2}\\)-database is a functor \\(H : \textrm{2} \rightarrow \textbf{Set}\\), ie an ordered pair of sets \\(\langle A, B \rangle\\).

There's a unique map \\(! : \textrm{2} \rightarrow \textrm{1}\\).

A \\(\textrm{1}\\)-database is a functor \\(F : \textrm{1} \rightarrow \textbf{Set}\\), ie a single set \\(X\\).

Given any \\(F\\) we can precompose with \\(!\\) to get a \\(\textrm{2}\\)-database \\(F \circ !\\).

This is the diagonal pair \\(\langle X, X \rangle\\).

The right Kan extension is the right adjoint to the functor \\(F \mapsto F \circ !\\).

It sends \\(\textrm{2}\\)-databases to \\(\textrm{1}\\)-databases, so the pair of sets \\(\langle A, B \rangle\\) is sent to a single set \\(P\\).

Given any \\(F\\), \\(H\\) we have \\(\textrm{Nat}[F \circ !, H] \cong \textrm{Nat}[F, \textrm{Ran}_!(H)]\\).

ie natural transformations from \\(F \circ !\\) to \\(H\\) correspond bijectively to natural transformations from \\(F\\) to \\(\textrm{Ran}_!(H)\\).

ie pairs of maps \\(a : X \rightarrow A\\), \\(b : X \rightarrow B\\) correspond to single maps \\(k : X \rightarrow P\\).

In particular we can set \\(F = \textrm{Ran}_!(H)\\), ie \\(X = P\\), and get a pair of maps \\(p : P \rightarrow A\\), \\(q : P \rightarrow B\\) corresponding to the identity \\(e : P \rightarrow P\\).

By naturality in \\(F\\) the composites \\(p \circ k : X \rightarrow P \rightarrow A\\) and \\(q \circ k : X \rightarrow P \rightarrow B\\) correspond to \\(e \circ k : X \rightarrow P \rightarrow P = k : X \rightarrow P\\).

And since the correspondence is bijective we must have \\(a = p \circ k\\) and \\(b = q \circ k\\).

Bijectivity also tells us that \\(p \circ k' = p \circ k\\) and \\(q \circ k' = q \circ k\\) implies \\(e \circ k' = e \circ k\\), ie \\(k' = k\\).

In other words, for any pair of maps \\(a : X \rightarrow A\\), \\(b : X \rightarrow B\\) there is a unique map \\(k : X \rightarrow P\\) such that \\(a = p \circ k\\) and \\(b = q \circ k\\).

ie \\(P\\) is product of \\(A\\) and \\(B\\), with projection maps \\(p\\) and \\(q\\).