No: the path of the free-falling observer at and beneath the horizon is not covered at all by the coordinate system of the external observer. That coordinate system simply becomes mathematically undefined, which means you can't use it to make any claims about what happens.

> In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike.

No: there is a separate coordinate patch that covers the region inside the horizon (but not the horizon itself--the horizon is not covered by either patch), in which the "areal radius" coordinate is timelike, and in which the coordinate corresponding to the extra Killing vector field is spacelike. Calling the former coordinate "radial" is at least justifiable, since it still is the "areal radius" coordinate even though it's timelike. But calling the latter coordinate "the time coordinate" is simply wrong; the fact that the letter t is commonly used for it does not mean it's a "time" coordinate in any meaningful sense.

> There is a coordinate singularity around this switch

Yes.

> which is where the infinite coordinate time comes from.

No, which is what makes statements like "infinite coordinate time" incorrect; the correct statement is that neither coordinate patch covers the horizon at all.

> An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon

Yes. The easiest way to do that is to switch to some other coordinate chart which covers the entire path, above, at, and below the horizon. In such a coordinate chart, the "areal radius" coordinate will not be timelike below the horizon. (In at least one chart commonly used for this, Painleve coordinates, all four coordinates are spacelike below the horizon.)

By "time coordinate" I mean the hyperbolic angle. This corresponds to the proper time of our constantly accelerating observer. This remains a well defined coordinate on the interior of the black hole. As you identify, in the language of relativity, this is a spatial coordinate on the interior of the horizon. I specifically refer to this as "coordinate time" because it is the coordinate that our observer is using for time; not because it is actually a time coordinate in all regions of space-time.

> No, which is what makes statements like "infinite coordinate time" incorrect; the correct statement is that neither coordinate patch covers the horizon at all.

You do not need to consider the horizon itself. Consider just portion of the path that occurs before the horizon. This path crosses infinite coordinate time. Put another way, the singularity in this coordinate system is precisely that coordinate time diverges towards infinity as you approach the horizon. Put another way, if you pick any finite amount of coordinate time, you can find a portion of free-falling path from before the horizon that crosses that amount of coordinate time .

[0] Although I think we need to be a bit more careful about what we mean by external observer for this to be formally true. A free falling observer is "external" until they cross the horizon, but they don't see a coordinate singularity there.

I mean that there are two separate, disjoint regions of spacetime (one outside the horizon and one inside the horizon), and each one is covered by a separate coordinate chart. Informally people often talk as though those two charts are "the same", but they're not. They're two separate coordinate charts, that just happen to be described using the same symbols.

> we seem to agree that an external observer sees a coordinate singularity at the horizon

No, we don't agree about that. The coordinate singularity is only there for the particular coordinates we are talking about (which cannot cover the horizon). There are plenty of other possible choices of coordinates that do not have a coordinate singularity at the horizon, so a single coordinate patch covers the entire region of spacetime of interest.

There is a certain sense in which the coordinates we have been using (Schwarzschild coordinates) are the "natural" ones for an observer who is at rest relative to the hole. But one has to be very careful not to put too much weight on that; it can easily lead to errors if taken too far.

> I tend to use a hyperbolic coordinate system.

Can you give a reference for this? What you are describing in your next paragraph (about "hyperbolic angle" and so forth) does not seem like any of the coordinate charts I am aware of for Schwarzschild spacetime (Schwarzschild, Painleve, Eddington-Finkelstein, Kruskal-Szekeres, or Penrose).

> In the region on the outside of the horizon, it matches the natural coordinate system for an observer maintaining constant acceleration away from the black hole.

This is just Schwarzschild coordinates; any observer who is at rest relative to the hole (a static observer) has to maintain constant outward proper acceleration to stay that way. Close enough to the horizon, these coordinates can be approximated by Rindler coordinates on flat spacetime, which are a commonly used coordinate chart in special relativity for treating observers with constant proper acceleration.

However, if Schwarzschild coordinates are what you have in mind, first, your next paragraph doesn't seem to be describing them at all, and second, the key property you give here does not hold on the patch inside the horizon, since there are no static observers there.

> Consider just portion of the path that occurs before the horizon. This path crosses infinite coordinate time

No, it doesn't. Every event on the path that is outside the horizon has a finite value of Schwarzschild coordinate time. You even say so later in this paragraph, so you appear to be contradicting yourself.

It is true that, for some cases, you can take limits as Schwarzschild coordinate time increases without bound to compute quantities that are finite on the horizon. In fact, the proper time experienced by an infalling observer from some altitude above the horizon, to the horizon itself, is one of them. But you cannot use this limiting process to justify saying that the Schwarzschild coordinate time itself "is infinite" at the horizon.

> A free falling observer is "external" until they cross the horizon, but they don't see a coordinate singularity there.

Any observer can choose any coordinates, regardless of their state of motion. Whether any observer "sees a coordinate singularity" at the horizon is purely a matter of which coordinates they choose, and has no physical meaning.