I assume here you are referring to the point that is stationary, i.e. the instantaneous centre of rotation.1. The disk rotates about its centre
2. The disk rotates about the point opposite to where the rope leaves the disk.
I prefer this argument. It is similar to the wheel on a car which, when viewed from a frame attached to the ground (as we have here), rotates about its point of contact with the ground. If there was friction then this is the intuitive answer suggesting that as I reduce friction to zero it will still rotate about that point.
3. The disk does not rotate because there is only one force acting on it.
4. The inertia of the disk provides "a restraint" which allows the single force to cause rotation. But, if so, how is the axis of rotation decided? Is it that about which the moment of inertia is the least - ie the centre of the disk?
It is hard to guess this in advance. As I posted, it is best to consider the disc's motion as the sum of a linear acceleration of the mass centre and an angular acceleration about that centre.
If the tension in the string is T then, that being the only force acting horizontally on the disc, the linear acceleration is a=T/M.
The torque the string exerts about the disc's centre is RT, so the angular acceleration is given by ##I\alpha=RT##. With ##I=\frac 12 MR^2##, we have ##M\alpha R=2T##.
We can get the corresponding velocities at time t by multiplying each by t.
Suppose the instantaneous centre of rotation is distance X from the disc's centre, distance X+R from where the string leaves the disc. We have ##X\alpha=a##, so ##2TX=MX\alpha R=MaR=TR##, whence X=R/2.