@alkaspeltzar I think you misunderstand what a torque is. I have the feeling this could go on forever, so I thought it might help if I spell out some things (but you should really read a textbook on intro classical mechanics).
You first have to choose a coordinate system; it doesn't matter if it's inertial or non-inertial, you just need to bear in mind that if it's non-inertial then you need to take into account inertial forces. Secondly, the
torque of a force with respect to this coordinate system is by definition ##\vec{\tau} = \vec{r} \times \vec{F}##, where ##\vec{r}## is to the
point of application. Thirdly, the vector sum of the torques on anybody is the rate of change of the angular momentum of that body, ##\sum \vec{\tau}_i = d\vec{L}/dt##; these quantities are all still calculated with respect to the same coordinate system.
The next important result is the König theorem, that the total angular momentum of a body of mass ##M = \sum m_i## can be written as the sum of the angular momentum that a particle of mass ##M## positioned at the centre of mass would have, plus the angular momentum of the body with respect to the centre of mass frame. That's because$$\begin{align*}\vec{L} = \sum_i m_i \vec{r}_i \times \dot{\vec{r}}_i &= \sum_i m_i (\vec{R} + \vec{r}'_i) \times (\dot{\vec{R}} + \dot{\vec{r}}_i') \\ &= M\vec{R} \times \dot{\vec{R}} + \left [\vec{R} \times \left(\sum_i m_i \dot{\vec{r}}'_i \right) \right] + \left[ \left(\sum_i m_i \vec{r}'_i \right) \times \dot{\vec{R}} \right] + \sum_i m_i \vec{r}'_i \times \dot{\vec{r}}_i' \\ &= M\vec{R} \times \dot{\vec{R}} + \vec{0} + \vec{0} + \sum_i m_i \vec{r}'_i \times \dot{\vec{r}}_i' = \vec{L}_{\text{CM}} + \vec{L}^*
\end{align*}$$In the general case, since ##\sum \vec{\tau} = d\vec{L}/dt = \frac{d}{dt} \left( \vec{L}_{\text{CM}} + \vec{L}^* \right)##, the resultant torque might cause changes in both the angular momentum about the centre of mass
and the angular momentum due to the translation of the centre of mass. (Remember, again, that ##\sum \vec{\tau}## and ##\vec{L}## are coordinate variant so we're still working with respect to one chosen coordinate system).
For a coordinate system whose origin is the centre of mass, the total torque (which here is also the total
real torque, because inertial forces have zero level arm) is just the rate of change of angular momentum about the centre of mass. It follows that if the lines of action of all of the forces acting on a body pass through its centre of mass, the angular momentum with respect to the centre of mass is constant. For the falling mass attached to the string, there is no resultant torque about its centre of mass, so the mass undergoes no rotation given that its initial angular velocity is zero.
The distinction is even clearer for something like e.g. a particle moving in the plane under the action of a force ##\vec{F}##. The angular momentum of the particle is solely due to translation of the particle w.r.t. the coordinate system, not due to any 'rotation of the particle'. You would find that ##\vec{\tau} = \vec{r} \times \vec{F} = \frac{d}{dt} \left( mr^2 \dot{\theta} \hat{z} \right) = \left( 2m r \dot{r} \dot{\theta} + mr^2 \ddot{\theta} \right) \hat{z}##. N.B. you might notice that if ##\dot{r} = 0##, then the RHS is just ##mr^2 \ddot{\theta} \hat{z} = I_z \alpha_z \hat{z}##; the angular acceleration ##\vec{\alpha}## in this instance is the angular acceleration of the line joining the particle and the origin.
