I think this is a very interesting topic but unfortunately the answer to "Has anyone written about it at length in the last 5 years?" is probably no. (It's possible that Dennis Gaitsgory has unpublished notes on it - you could try emailing him to ask.) Are there specific parts of "How to invent shtukas" that you would like elaborated on? I can try to write an answer fleshing out some of the details; it would be helpful for doing so to know which parts to focus on.

@TabesBridges Sorry, "birational" and "biregular" in my previous comment should have been "rational" and "regular."

@user127776 Considering the varieties up to birational equivalence often doesn't do anything because of the following fact: If $f:X\rightarrow Y$ is a birational morphism of smooth projective varieties and $Y$ contains no rational curves, then in fact $f$ is a literal (biregular) morphism.

@user127776 It is true that my argument does not apply to that case. More generally though, statements of the form "every variety has a nice finite cover" tend to be false, which is unfortunate because they would be extremely useful. P.S. the case of products of curves has some interesting history behind it: see mathoverflow.net/questions/98771/…

Re question 3: The answer should be no for any formalization of this question (basically because by passing to finite covers you generally cannot make the variety less complicated). For instance, for every family of varieties, a very general hypersurface of large degree will admit no birational morphisms from any member of that family.

Let us continue this discussion in chat.

@derryberry BTW, I think what your argument shows is the following. Instead of working on $\Delta_3^{(1)}$, let $Y$ be the fiber product $\Delta^{(1)}\times_{X}\Delta^{(1)}$. This contains but is strictly bigger than $\Delta_3^{(1)}.$ Its projection to the first and third factors lands in $\Delta^{(2)}$ instead of $\Delta^{(1)}$, so we can define $\phi_{12}$ and $\phi_{23}$ on it but not $\phi_{13}$. Then, if $\phi_{23}\circ\phi_{12}$ is the pullback of a morphism on $\Delta^{(2)}$ whose restriction to $\Delta^{(1)}$ equals $\phi_{13}$, the connection on $\mathcal{E}$ is flat.

@derryberry I'm not sure what the best reference is, but I think people.math.harvard.edu/~gaitsgde/grad_2009/SeminarNotes/… does it in some detail. Maybe it's in Grothendieck's "Crystals and the de Rham cohomology of schemes?" One way to do it is something the following: First show that the cocycle condition corresponds to compatibility with some multiplication on $\mathcal{D}$. Then use the fact that $\mathcal{O}$ is a $\mathcal{D}$-module to conclude that it must be the standard multiplication.

@derryberry I do agree that if you add a cocycle condition to the formal lifting property then all is good. But I'm still skeptical of the argument for integrability: The issue is that looking at the first order neighborhood is not enough to "see" $\mathcal{D}^{\leq 2}$. I think the cocycle condition that you're trying to impose for integerability translates not to what you want it to, but instead to additivity of the $\mathcal{D}^{\leq 1}$ action, which is automatic.

When $r$ and $d$ are relatively prime, I think this follows from rationalality of the fibers of $M(r,d)\rightarrow\operatorname{Jac}^d(X)$ (which I think is proven in arxiv.org/abs/math/9907068). In that case the map to $\operatorname{Jac}^d(X)$ is the Albanese map of $M(r,d)$ and so if the $M(r,d)$ are the same so are the $M(r,\xi)$. I am not sure what happens when $r$ and $d$ are not relatively prime.

@Meow I am now very confused. Are you working in the algebraic or the analytic world? If you are working in the analytic world then I agree with what you are saying, but then I don't know how to make sense of even the normal de Rham stack $X_{dR}$. If you are working in the algebraic world then I see no difference between meromorphic connections and $D$-modules on $U$. (In particular, $\mathcal{O}_X(D)$ and $\mathcal{O}(U)$ coincide in the algebraic world no?)

@Meow Why is it not?