I'm only familiar with topic from pictures like this; here's Schwarzschild coordinates are gradually morphing into Finkelstein (source .gif):
My question is about light rays on pictures 1-3: do they cross the horizon? Or they're asymptotes?
It's not clear on pictures 1-3, but I'm sure it's clear in math (which I'm needing a tip with).
Also, it's obvious on a picture 4 that they're crossing horizon, but I'm asking anyways.
The light rays always cross the horizon. You just can't see it in Schwarzschild coordinates because the coordinates themselves don't cover the horizon. Taking the black hole radius to be $1$ like your source does, $r_\text{F}=1$ in Eddington-Finkelstein coordinates is the horizon, but $r=1$ in Schwarzschild coordinates is just a coordinate singularity. When morphing between the coordinate systems, at some stage you have to replace the fake horizon with the real one, and that's when the light rays will appear to start crossing it.
At all radii except $1$, the Schwarzschild and Eddington-Finkelstein time coordinates are related by $t_\text{F}=t-\ln\,\left|r-1\right|$ (with the $r,θ,\phi$ coordinates being the same for both). An obvious way to interpolate between them is to plot $t_k=t-k\ln\,\left|r-1\right|$ for $k$ ranging from $0$ to $1$. If you do that, and also define the points at $r=1$ by a limit, then you'll find that the real horizon only appears at $k=1$. At all smaller $k$ you just have a singularity. So the light rays will appear to suddenly start crossing the horizon at the exact same moment that the coordinates become Eddington-Finkelstein. This is most likely how it was done in the image in the question.
Ari said:
Schwarzschild co-ordinate (SC) is associated with an asymptotic observer, someone who is stationary at infinite distance from the black hole.
This is a frequently made claim, but it's just false. First, there's no such thing as a stationary observer at infinity in an asymptotically Minkowskian spacetime. You can see that by looking at the Penrose diagram. Spatial infinity is the single point at the far right. You can't be at rest there because there's no time direction to move in.
Second, there's no such thing as the coordinate system of an observer in general relativity – unless you define "observer" to mean "coordinate system", as is often done in special relativity, but then the observer doesn't have a position. Any observer can use any coordinate system. Even if you demand that they use a system whose coordinate time matches their proper time (and I don't see why you should), that still leaves an infinite, and extremely varied, choice of coordinate systems.
In this co-ordinate system the light rays asymptote at the event horizon. This can be interpreted as infinite redshift. [...] [E-F] is non-singular at the horizon, so there's no infinite redshift and the light rays can go in.
This is also wrong. There is, physically, an infinite redshift of objects crossing the horizon as seen from outside. This is a prediction of the theory and as such it's independent of coordinates. The fact that ingoing light rays cross the horizon is also a physical fact.
Whether a physical phenomenon happens can never depend on an arbitrary choice of coordinates.
The co-ordinate systems mentioned in your question are associated to different observers, hence their characteristic is different.
Schwarzschild co-ordinate (SC) is associated with an asymptotic observer, someone who is stationary at infinite distance from the black hole. In this co-ordinate system the light rays asymptote at the event horizon. This can be interpreted as infinite redshift. In other words, it seems to this observer that any free-falling particle takes infinite amount of time to reach the horizons and they never cross it.
The (ingoing) Eddington-Finkelstein co-ordinate (EF) is associated with null geodesics that are entering the BH, so basically the path of photons that are going in. This metric is non-singular at the horizon, so there's no infinite redshift and the light rays can go in.
I'm not sure what morphing of two co-ordinate system means in your question. But, if you start as a asymptotically stationary observer (SC) and then then gently nudge yourself towards the Black Hole geometry into a free fall, your co-ordinate system will look qualitatively like EF (not entirely, since you won't be following a null geodesic). Hence, at first you'll see light rays asymptote at the event horizon but during your fall you'll observe them going in.
