Interpretation of this spacetime diagram for freefalling observers in Gullstrand-Painleve coordinates? The following diagram (which I've adapted from Chapter 7 of "Exploring Black Holes" by Taylor, Wheeler & Bertschinger) shows the radial world-line of a free-falling observer into a Schwarzschild black hole (labelled as "world line of a raindrop") and the null geodesics followed by light emitted by that observer, both radially inwards and radially outwards (labelled as ingoing and outgoing flashes). The diagram uses Gullstrand-Painleve coordinates $r, T$.
I have added a further (red) worldline, which represents the trajectory of another free-falling observer (let us call them observer 2) that crosses the event horizon (at $r/M=2$) at a larger value of $T$.  Observer 2 is capable of receiving light signals from the first observer up to point E, at which point, the outwardly directed null geodesic intersects with the world line of observer 2 at point G, at $r=0$.

So far so good. But imagine, rather than just the first observer drawn, there were a series of freefalling light beacons that crossed the event horizon prior to observer 2 at a range of $T$ values, stretching into the past.
Questions:
Once observer 2 has crossed the event horizon, is it correct that any light they receive from a beacon that has already crossed the horizon would appear to come from behind them? (Where behind means from the direction of increasing $r$).
Up until they reach the event horizon, the light received by observer 2 from all the beacons would come from the direction of decreasing $r$. So does that mean that as observer 2 crosses the event horizon they essentially "pass through" the images of all the beacons at once? And in fact, must they "pass through" the outward facing images of everything else that has previously fallen into the black hole? What on Earth would that look like?
Finally, what would be the Doppler shifts of the light received by observer 2 in these cases?
 A: The angle at which a lightlike worldline crosses observer 2's worldline in the diagram should not be interpreted as the direction in which obsever 2 must look to see the light.
Here's one way to understand it. Consider two infalling beacons, one just ahead of observer 2 and one just behind observer 2. At any given point along observer 2's worldline, the diagram will show two lightlike worldlines crossing that point: one from one beacon, and one from the other. Both of these lines will be angled to the left in the $r/M<2$ part of diagram, but one will have a steeper slope than observer 2's worldline, and one will have a shallower slope. The relative slopes tell us which way observer 2 needs to look in order to see that light.
This is consistent with the equivalence principle: in a sufficiently small neighborhood of any given observer, spacetime is essentially flat. This is true at every point in the Gllstrand-Painlevé diagram (except at the curvature singularity, where "sufficiently small" collapses to an empty set). The fact that all causal worldlines are angled to the left in the $r/M<2$ part of the diagram is just an artifact of trying to depict the whole curved spacetime in a flat two-dimensional space. We could draw Minkowski space in a similarly distorted way, if we wanted to. The appearance of the diagram depends on which coordinate system we use, and no matter what coordinate system we use, the diagram cannot faithfully represent the geometry — partly because the geometry is curved, and partly because it has Lorentzian signature.
To answer the first question directly: If a beacon crosses the horizon before observer 2 does, then observer 2 will continue looking forward to see the light from that beacon. If a beacon crosses the horizon after observer 2 does, then observer 2 will continue looking behind to see the light from that beacon. The observer never passes through the image of any of the beacons, not even when the observer hits the singularity. In fact, the observer never sees the leading beacon hit the singularity: the last thing the observer sees is some of the light that the beacon emitted before it hit the singularity. This is consistent with the fact that the singularity isn't a point in space. It's an event (I mean, a locus of events) in the future, and that remains true no matter what coordinate system we use and no matter how we draw the diagram.
The question about Doppler shifts requires some calculation. It depends on how far apart the observer and beacon are. Qualitatively, if the beacon falls in just ahead of the observer, say initially 1 meter away, then the observer won't notice any significant redshift when crossing the horizon (if the black hole is big enough). As they approach the singularity, I expect that the observer would see an increasing amount of redshift due to the differential curvature of spacetime (tidal effects) between the beacon and the observer, but I haven't done the calculation to quantify this.
A: As an addendum to Chiral Anomaly's excellent answer, I want to consider the second part of the question:

Up until they reach the event horizon, the light received by observer 2 from all the beacons would come from the direction of decreasing r. So does that mean that as observer 2 crosses the event horizon they essentially "pass through" the images of all the beacons at once? And in fact, must they "pass through" the outward facing images of everything else that has previously fallen into the black hole? What on Earth would that look like?

This actually not any different then the following setup in flat Minkowski space. Suppose our observer sits at the origin, and we have a series of stationary beacons setup along a straight line from the observer. If we follow the past light cone of the observer, it will intersect with all the worldlines of the beacons. So when looking along the line of beacons, you see the superimposed image of all beacons (or more realistically just the first beacon blocking the line of sight to all others). What on Earth would that look like? Well, just look out of the window.
The situation with the series of beacons falling into a black hole is no different. For concreteness sake, lets suppose that the beacons and the observer are all following the same radial trajectory, but shifted by a constant amount along the Killing vector field that is timelike at infinity, i.e. the beacons and the observer have been dropped from the same position at infinity at regular intervals. At any point during the observers one way trip to the singularity, when he looks forward --- i.e. along his past lightcone in the forward direction ---  he will see the superimposed images of all beacons. In particular, this is also true when he passes the event horizon (and after, all they way to the singularity). In this regard nothing special happens at the event horizon.
In this setup, we can also easily answer the question about the redshift of the beacons as the observer crosses the event horizon. This is easy because the null ray along the event horizon that connects the observer to all the beacons is actually generated by the Killing vector field used to separate the trajectories. Consequently, the four velocity of the observer parallel propagated (back) along this ray will always be equal to each of the four velocities of the beacons as the ray cross them. Consequently, there will be no observable redshift of any of the images of the beacons that the observers see at the moment he crosses the event horizon.
A: A couple of points not covered in answers by ChiralAnomaly and mmeent:

*

*Since before observer crosses the horizon the light from previously fallen beacons would appear redshifted and at the horizon the redshift is zero, after the horizon crossing the light from previously fallen beacons would appear blueshifted (but coming from the same direction) to the observer.


*At the moment of horizon crossing, the energy flux that reaches the observer from beacons that have fallen long time before ($T_\text{b}-T_\text{o}\gg M$) would be exponentially small. This could be seen as a consequence of continuous exponential stretching of space in radial direction that happens along the null horizon generators, so receiving even a single photon from a beacon fallen long before is an extremely unlikely event.
