I think that the distinction between active and passive is perhaps clearer when you use a simpler transformation than a full diffeomorphism. So let's take invariance under Galilean transformations:Wrt to the passive vs active diffeomorphism issue I can't see how it can have any physical consequence, it is just a way to allude to the fact that every diffeomorphism is bijective and therefore every active (passive)transformation that is a diffeomorphism has its inverse passive (active) transformation
x' = x - vt
t' = t
There is an active and a passive interpretation of this transformation. The passive interpretation is about the form of the equations of motion: If you start with the equations expressed in terms of (x,t), and then you rewrite the equations in terms of (x',t'), then the equations have the same form. For example, if the equations of motion are:
m d2x/dt2 = F
then since d2x/dt2 = d2x'/dt'2, the equations have the same form in terms of (x',t'), provided that the force F transforms as:
F' = F.
On the other hand, the active interpretation is about solutions to the equations of motion: If x = f(t) is a solution to the equations of motion, then x = f(t) - vt is also a solution to the equations of motion.
Notice the distinction: In the passive case, we are considering the same situation expressed in different coordinates. In the active case, we are consider a different situation expressed in the same coordinates. Newton's equations, with a velocity-independent force, are invariant under both types of transformations.
But now let's spoil the invariance by throwing in a friction force proportional to the velocity. We can still write the equations in a way that is invariant under passive transformations:
m d2x/dt2 = F - k (dx/dt - u)
This is still invariant in form under a Galilean transform, provided that we transform things as follows:
x' = x - vt
t' = t
F' = F
u' = u - v
So things are still invariant under passive transforms. But what about active transforms?
If x = f(t) is a solution to the equations of motion, is x = f(t) - vt also a solution to the equations of motion? Clearly not.