This is going to be a long answer because there is a lot to unpack here.
I'm going to briefly go over your questions about your understanding, then analyze the experiment from the video in detail, then explain why I think this experiment is not very interesting (since the video also tries to hype it).
Your understanding
if the information in the detectors is erased after detection by the detector, but before any observation by the experimenter (of either the detector or the screen), then we get an interference pattern, just as if the detectors were not there.
It doesn't matter when the "eraser" is used, or when the experimenter looks at anything; you get the same results regardless. Also, there's never an interference pattern visible on the screen. That pattern only shows up in later data analysis.
whether we get an interference pattern depends on whether the "which-way" information "exists" (at the time of observation by the experimenter).
That's correct in a certain sense. The way quantum mechanics works is that the system (as encoded in the wave function) interferes with itself as a whole. If there's which-path information anywhere in the system, then the paths are orthogonal, so they don't interfere.
But note that applying the "eraser" doesn't destroy the which-path information for this purpose, nor does any measurement, or anything else you can do without involving the photon. There is no way to destroy the which-path information unless you re-combine it with the photon, which doesn't happen in this experiment. "Quantum eraser" is really a misnomer.
detection by the detectors, in itself, does not collapse the wave function
That's correct of the "detectors" in this experiment. Typical thought-experiment "detectors" do collapse the wave function, though.
observation (and perhaps other actions taken) by the experimenter is different in some way to detection by the detector.
Observation by the experimenter is thermodynamically irreversible (and collapses the wave function), while the "detection" (by these detectors) is reversible (and doesn't). That distinction isn't important in this experiment, though. It's important that the "detection" at the slits be reversible, but the experimenter and the screen could just as well be reversible quantum computers. But it's probably easier, and just as correct, to think of them as wavefunction-collapsing classical objects.
The video (hype)
The video has some silly hype at the beginning:
[The experiment] demonstrates that the consequences of quantum mechanics are far stranger than we had ever imagined,
I disagree with this; I think that it's just a less interesting version of the EPR/Bell/Aspect experiment. I'll explain what I mean by this after the analysis.
and some people even claim that it indicates that messages are being sent backwards in time.
Those people are wrong. This isn't a matter of interpretation; they just don't understand how probability works. I'll come back to this after the analysis.
The video (post-hype)
I think (contra David Reishi) that the video is pretty accurate when it sticks to the physics, once you figure out what it means by "detector" and "quantum eraser".
In the video, the "detector" at each slit contains a two-state quantum system (a qubit) which is initially in some known state (say $|0\rangle$) and is flipped (to $|1\rangle$) if the photon passes through the detector. It might be quite hard to engineer such a device, but the laws of physics allow it: it's just a CNOT gate with the presence/absence of the photon as the control bit. (Arguably this shouldn't be called a detector or a measurement because it is reversible, but those are just words; it's clear what the device does physically.)
I'm going to simplify the setup a (qu)bit by omitting one of the detectors, because that doesn't lose any information: if the photon didn't go through that slit then it went through the other (in this idealized experiment free of engineering realities). The single detector is on the right slit, and gives us one qubit, which is $|0\rangle$ if the photon went through the left slit and $|1\rangle$ if it went through the right slit.
The "quantum eraser", which looks like a reject from a new-age crystal-healing video, is a quantum computer which simply applies a Hadamard gate to the qubit (that is, it takes $|0\rangle$ to $(|0\rangle+|1\rangle)/\sqrt2$ and $|1\rangle$ to $(|0\rangle-|1\rangle)/\sqrt2$).
The most serious error, or misleading statement, in the video is the implication that the "quantum eraser" erases which-path information, allowing the wave function to interfere again (13:04). That is not possible in quantum mechanics. There is nothing you can do to the qubit (or anything in the universe other than the photon), at any time, that will affect the observable behavior of the photon.
Analysis
Just after the "detection", the system is in the state $$|0\rangle|\text{photon in left slit}\rangle + |1\rangle|\text{photon in right slit}\rangle$$ (times $1/\sqrt2$; I'm going to ignore normalization factors for the most part). The photon then propagates to the screen. Just before the photon hits the screen, the state of the system is $$\sum_P (\alpha_P |0\rangle|\text{photon at P}\rangle + \beta_P |1\rangle|\text{photon at P}\rangle) = \sum_P (\alpha_P |0\rangle + \beta_P |1\rangle) |\text{photon at P}\rangle$$ where the sum is over all the points on the screen (all the pixels of the CCD, if you like).
If we had not put a detector at the slit, that sum would have been $$\sum_P (\alpha_P |\text{photon at P}\rangle + \beta_P |\text{photon at P}\rangle) = \sum_P (\alpha_P + \beta_P) |\text{photon at P}\rangle,$$ and the probability of finding the photon at P would therefore have been $|\alpha_P + \beta_P|^2$, which depends on the relative phase of $\alpha_P$ and $\beta_P$ (whose phases depend on the distances to the left and right slits respectively). But with the detector at the slit, the states scaled by $\alpha_P$ and $\beta_P$ are orthogonal, and so the probability of finding the photon at P is $|\alpha_P|^2 + |\beta_P|^2$, which does not depend on the relative phase. This is why the interference pattern disappears. In effect, there is interference only between identical states of the whole world, not just identical states of the photon we're detecting.
Suppose the photon is actually detected at P (which is now a specific point, not a bound variable of the summation). The state of the system (after wavefunction collapse) is now $\alpha_P |0\rangle + \beta_P |1\rangle$ times an overall normalization factor that we don't care about.
At this point we have a choice (the delayed choice): measure the qubit in the {0,1} basis, or apply the "quantum eraser" (Hadamard gate) and then measure the qubit in the {0,1} basis. In the former case, we'll get 0 and 1 with relative probabilities $|\alpha_P|^2$ and $|\beta_P|^2$ respectively. In the latter case, we'll get 0 and 1 with relative probabilities $|\alpha_P+\beta_P|^2$ and $|\alpha_P-\beta_P|^2$ respectively. (They're relative because of the normalization factors that I ignored.)
Now think about what happens if we condition (in the ordinary classical sense) on getting 0 or 1 in this measurement. If we didn't apply the Hadamard gate, $|\alpha_P|$ and $|\beta_P|$ just fall off monotonically with increasing distance from the respective slits, and they will be roughly equal, so we'll get 0 for around half the dots and 1 for the other half, across the board. If we did apply the Hadamard gate, $|\alpha_P \pm \beta_P|$ oscillate at the frequency of the interference pattern we would have gotten in the standard double-slit experiment. If P is near one of the peaks of $|\alpha_P+\beta_P|$, we'll almost certainly get 0; if it's near one of the troughs, we'll almost certainly get 1. So conditioning on Hadamard 0 or 1, we'll get an interference pattern.
Why the people who think this shows backward causation are wrong
You have a bag containing 4 balls, 2 red and 2 black. You draw a ball. There's a 1/2 chance it will be red. If it is red, there's a 1/3 chance that the second ball you draw will be red.
But if you don't look at the first ball, there's a 1/2 chance that the second ball you draw will be red, and if it is, there's a 1/3 chance that the first ball you drew was red. If you collect data over many trials, conditioned on the second ball being red, you'll find that indeed about 1/3 of the first balls you drew were red.
Backward causation? Of course not. If X is correlated with Y, then Y is correlated with X. It doesn't matter whether Y happened after X.
The argument for backward causation in the delayed-choice quantum eraser experiment is exactly the same as the argument for backward causation in this classical experiment.
Why this is a less interesting version of Bell's experiment
The initial state ($|0\rangle|\text{photon in left slit}\rangle + |1\rangle|\text{photon in right slit}\rangle$) is just a Bell state ($|0_A\rangle|0_B\rangle + |1_A\rangle|1_B\rangle$). There are three major differences between this experiment and the Bell experiment:
The measurement axes range over the whole Bloch sphere (linear/elliptical/circular polarization for photons, or any spin axis for electrons), while in the Bell experiment they're restricted to a plane (linear polarization for photons, or a spin axis in some plane for electrons).
The universe chooses the orientation of one of the detectors. (The dot on the screen encodes both the orientation and the result of the measurement, in this analogy.) For the other measurement we choose between two orientations (eraser or no eraser).
The measurements of the two halves of the Bell pair are timelike separated, not spacelike separated.
The first difference doesn't matter much; I think you can still derive a version of the Bell inequality with this change. (If you can't, that's just another reason why this experiment is less interesting.)
The second and third differences make the experiment much less interesting, because they each independently make the whole thing consistent with a local hidden variable theory.
You can think of Bell's experiment as a game show similar to The Newlywed Game: the contestants are allowed to talk and agree on a strategy, then they're separated and independently asked questions that they didn't know in advance. Their goal is to give answers that are correlated in a certain sense.
If they're allowed to choose the question asked of contestant A, they can win easily, since they can agree on contestant A's question and answer in advance, and contestant B then has all relevant information when deciding how to answer. Likewise, if B's question is in the future light cone of A's question and answer, they can win if A surreptitiously sends that information to B by radio or dark matter or other light-speed-limited means.
You could replace my ball experiment with a classical simulation of the delayed-choice quantum eraser. For example: a computer generates two random numbers $x,y \in [0,1]$ and prints out two copies which are given to two oracles. The first oracle simply tells you the value of $x$. The second oracle, on your command, either tells you whether $y < \tfrac12$ or whether $y < \sin^2 100x$. If you condition on the answer to the first question, you get a flat histogram of $x$ values, while if you condition on the answer to the second question, you get a sinusoidal histogram. With minor tweaks to the second oracle's formulas, this simulation becomes exact.
Bell wrote an essay called "Bertlmann's Socks and the Nature of Reality" in which he presented a thought experiment similar to my four-ball experiment (involving his colleague Dr. Bertlmann who always wears mismatched socks), as an example of what the EPR experiment is not about. I think the people who invented the delayed-choice quantum eraser didn't understand the difference between quantum mechanics and Bertlmann's socks.
