An (undriven) damped harmonic oscillator (satisfying $m\ddot{x} + \gamma \dot{x} + \omega_0^2 x=0$) can be solved by the solution(s) $x_0e^{i \omega t}$. For an underdamped oscillator these solutions represent pure oscillations mixed with exponential decay(/growth). Because both solutions for $\omega$ oscillate with the same period, all combinations of them also oscillate with the same period.
I suppose that your confusion arises from an intuitive idea of why a restoring force leads to periodic motion. In the undamped case the phase space trajectories of the particle are closed (i.e. the particle always returns to the same position(s)). In the damped case the trajectories spiral towards rest. But both of these motions are periodic in the sense that they reach their relative extrema (and zero crossing) after a specific interval of time. How can you see this? Perhaps you can take some inspiration from the undamped case: notice that the period is independent of the size of the orbit. Because the equation is linear any change in amplitude is exactly accounted for by a change in the force. In fact all linear (homogenous) ODE's can be satisfied by the anstaz above (i.e. solutions are periodic, damped, or both).
$v = |\vec{\omega}\times\vec{r}|= \omega r$ is always valid for a rigid rotating body. Here, $r$ refers to the distance of any particular point from a chosen axis of rotation, $\omega$, the angular speed of the body about that chosen axis and $v$, the linear speed of that point perpendicular to the radius vector (or the line joining the axis to that point).
In the image given below, for example you could say
$$V_\perp =|\vec{\omega}\times\vec{r}|= \omega r = r \frac{d\phi}{dt}$$
So it is, in general, a relation between the angular speed and linear speed of a point. This is always true.
However, in the problem something else has been said. In the problem, $v$ refers to the velocity of the center of mass of the ring. So, the problem just says that the center of mass of the body is moving at some particular speed $$v_{cm} = \frac{\omega R}{2}$$
There is nothing wrong here, the center of mass can move at any speed. Now, if you chose the center of mass as the center of rotation, then you have a rotation at $\omega$ superimposed on a translation at $v_{cm}$. Any rolling motion can be decomposed into a pure rotation and a pure translation, as shown below. Then the two effects can be superimposed.
About this center of mass, $$v_{rot}=\omega r$$ So the point in contact with the ground (at a distance R from the C.M.) for instance, has linear speed $v_{cm} - R\omega$ since rotation and translation oppose each other at that point. As others have pointed out this is not 0 and that implies there is some relative motion between the ring and the ground at the point of contact. This is called slipping.
So your friend was correct. Both are valid and correct. But they refer to different things - one to the motion of the center of mass and the other to that of any point on the body.
Best Answer
Angular frequency corresponds to the rate at which an angle is changing, so you will most likely find it as part of the argument of trig functions/complex exponentials. It is usually understood like "radians per time"
"Regular" frequency is usually anything else, typically like a "number of events per time". This could be talking about number of rotations, cycles, or anything really. As long as it has a will defined period between the same events.
In either case, the terminology is not as essential as knowing what you are denoting as $f$ or $\omega$ really means in the system you are looking at. They are usually interchangeable anyway through simple conversions.