You are correct that the angle between Q and P is not the same as the angle between Q' and P. However, the length of Q is also not the same as the length of Q'. The cross product depends on both of these things. It might not be immediately obvious why these two changes exactly compensate for one...
In Newtonian mechanics, F=ma will hold. Actually, that's tautological because Newtonian mechanics is defined by Newton's laws of motion and F=ma is Newton's second law.
Anyway, often the tricky part is to figure out what F is. In the E-frame, or any other non-inertial frame, you just have to...
All the possible fictitious forces pertain to this problem, which is partly what makes going to the E-frame difficult. There is a linear fictitious force parallel to the incline just because the disk as a whole is accelerating down the incline. There are also the three rotational fictitious...
Well, the unit vectors of a frame can always be made constant with respect to an observer in that frame so their constancy alone can't determine whether or not the frame is inertial. I won't bore you with inertial reference frames since that's probably on wikipedia or something and the problem...
So, it's all a matter of where you define your origin to be. As drawn, \vec{r} is with respect to the center of the disk. If the center of the disk is the origin, then \vec{r} = \frac{R}{2} \hat{e}_r.
If you say \vec{r} = R \hat{n}_2 + \frac{R}{2} \hat{e}_r, then you have defined your origin...