My apologies, I won't be reading your entire question.
But still I will provide an answer. Why is that? Because flight does not require any of the things you talk about.
You could build an airplane that would fly with no "airfoil" shapes. You could build an airplane that would fly with completely flat rectangular wings made out of plywood. The important thing would be the angle of attack of the wings to the air. Consider a flat piece of wood, like plywood. Push it through the air in a direction exactly parallel to its flat dimensions and it develops no lift. Tilt the wood so the leading edge is "up" compared to the direction it is moving and you can feel the lift.
The lift can be thought of a few ways. Think of the air molecules hitting the surface of the wood. They bounce off, in a downward direction. Well if we are pushing air downward, we must have an equal and opposite force, which is the lift. Or another way: we are gathering air under the board, it gets a little pressurized. The pressure is pushing up on the wood. This is really the same picture as the first if you think about it.
All the rest with airfoils and so on, all this has to do with developing lift efficiently, developing lift while minimizing drag. An airplane with flat plywood wings would fly, but it would have a lot of drag and would therefore be very inefficient.
From my understanding, what's happeneing is the adverse streamwise
pressure gradient precludes the boundary layer from progressing
downstream past a certain point, and the upstream flow subsequently
has nowhere to go but up and off of the body.
This is correct, in a sense. The effect of an adverse pressure gradient is to decelerate the flow near the body surface. This can be seen, for example, by examining the boundary layer equation in two dimensions.
$$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}=\nu\frac{\partial^2 u}{\partial y^2}-\frac{1}{\rho}\frac{\partial p}{\partial x}$$
If you consider steady flow and assume normal velocities to be small, then by inspection, we can see that an adverse pressure gradient causes $u$ to decrease in the streamwise ($x$) direction.
As you suspected, separation requires that the flow near the boundary stagnates. Moreover, separation occurs when the flow actually reverses.
$$ \frac{\partial u}{\partial y}_{y=0}=0; \quad \text{Flow Stagnation / Impending Reversal} $$
Additionally, it requires that the pressure gradient be simultaneously adverse, so that the the flow does not accelerate again.
$$ \frac{\partial p}{\partial x}>0 \quad \text{Adverse Pressure Gradient}$$
So, in short, you're correct. However...
This is a very different causal relationship from the first
explanation, where the flow lacks a sufficient streamwise-normal
pressure gradient to overcome the centrifugal forces of a curved
streamline.
The two statements are essentially the same - there are any number of ways to physically describe what's going on- but I think you've got the causality mixed between the two. The curvature of a body, and thus its attending streamlines, jacks up the adversity of the pressure gradient along that body (assuming you're past the point of minimum pressure). So it's the adverse pressure gradient that ultimately leads to separation. In a perfect world, where viscosity didn't exist, the flow would speed up as it hits the forward part of a curved body. The pressure would drop as it reaches the widest point of the body, streamlines are "squeezed" together, and the flow reaches a maximum velocity. On the afterbody, the flow would decelerate and the pressure would increase until both reach their upstream values. It's a simple trade between kinetic energy (velocity) and potential energy (pressure). In a real viscous flow, some of that kinetic energy is dissipated in the heat-generating nuisance that is a boundary layer, so that when the transfer from kinetic back to potential energy occurs on the afterbody of a curved surface, there isn't enough kinetic energy, the flow stagnates and reverses, and you get flow separation.
I can't comment on shock-induced separation, as I work in hydrodynamics and don't worry about compressibility. I'm no authority in that area, either, so if somebody takes issue with my explanation, feel free to criticize.
Best Answer
The reason is to keep velocity gradients small. This in turn means that some of the fluid medium which is trapped in a cavity or a region of separated flow will be moved in the direction of the outer flow. To counter this transport of trapped medium, a reverse movement at the opposite side of that region is needed to maintain uniform density.