Presumably it's a snapshot of X at a single instant of time, the initial state of the thought experiment. The author should be clearer however.

I agree, if this is the author's intent it should be clearer. However, I'm willing to interpret the diagram that way for this discussion.

GR allows the escaping particles to never be overtaken by the horizon even though GR also demands that those particles move slower than the horizon does.

Because spacetime is curved, and the curvature limits the size of a local inertial frame straddling the horizon.

Consider a "cloud" particle in the right half of cloud in the Frame X diagram. In order to be moving at escape velocity at its location, it must be moving to the right at just a smidgen less than c. So it will move off the right edge of the diagram, and out of the region covered by Frame X, before the horizon catches it. Once it's out of Frame X, its worldline doesn't have to look like a straight line in Frame X, so it can continue to avoid the horizon; in fact it will gradually get farther and farther away from it.

Another way to see what's going on is to ask what a line of constant radius r looks like in Frame X. An observer at constant r close to the horizon has to accelerate very hard to stay at the same radius; that means a line of constant r in Frame X is going to look like a hyperbola in a spacetime diagram. If we imagine a bunch of observers at constant r passing through the cloud in the Frame X diagram as shown, they will all be accelerating very hard to the right, hard enough to keep the horizon from catching up with them; but a cloud particle moving at escape velocity just a little to the right of the horizon will be moving to the right faster than any of the constant-r observers, so he will pass them one by one as they accelerate; they won't be able to catch him. That means the escape velocity particle is increasing its radius with time, i.e., escaping.

If all the particles above the horizon are escaping then the cloud is splitting along the horizon.

Because the particles that are escaping have to be moving to the right at very, very close to the speed of light, whereas the particles that are not escaping are not. This is not a contradiction.

[Edit: I suppose the intent could be that all of the cloud particles are moving to the right at just a smidgen less than c; but if that's the case, the ones to the right of the horizon line will make it out of Frame X before the horizon catches them, as above, but the ones at or to the left of the horizon line obviously won't, because the horizon will move ahead of them.]

You can't say it has no location in space and then refer to its location in space even in relative terms like "in the center of".

I haven't done any such thing. You appear to be reading things into my posts that I am not saying. I have said, consistently, that the singularity is a moment in time. More precisely, the singularity is represented by a spacelike line. A "place in space" is represented by a timelike line, not a spacelike one. A point in space at an instant of time is just a point, not a line.

The singularity has a location in both space and time

No, it has a location in time only; it is all of space (or at least all of it inside the horizon) at that moment in time.

I suspect you're getting your idea from a cosmic singularity, like for the big bang.

The black hole singularity is indeed very similar to the big bang singularity; both of them are spacelike lines. The only real differences are that the black hole singularity is hidden behind a horizon, and that the black hole singularity is a future singularity while the big bang singularity is a past one.

You say you have taken classes on this material; have you raised the issues you are raising here in class? If so, what response did you get?