(You can skip this derivation and go down to my final question if you already are familiar with the results $(1)$ and $(2)'$ from this derivation)
Suppose we are in the xy-plane:
In two dimensions, you can define the torque $\tau$ as $$\tau = I_O \ddot{\theta}$$ where $I_O$ is the moment of inertia of a body around the origin $O$ and $\ddot{\theta}$ is the angular acceleration of the bodys center of mass around the origin.
According to the parallel axis theorem, the moment of inertia of a particle around the origin is $$I_O = d^2M$$ where $d$ is the distance between the origin and the particle and $M$ is the mass of the particle.
Now let's suppose that the position in polar coordinates of a particle of mass $M$, referenced from the origin, is given by the vector $$\mathbf{r} = r\hat r+\theta\hat\theta$$
The distance to the origin is simply $r$ so we get $$I_O = r^2M$$
so the torque is simply $$ (1):= \tau = Mr^2\ddot{\theta} $$
We calculate the torque again by using the known fact that $$(2):=\tau = Force \cdot Distance = Fr$$ where $F$ is the net force on the particle in the $\hat \theta$ direction, since the force component in the $\hat r$ direction doesn't contribute to the torque.
According the newtons second law, we can write the force $F$ as the product of the particles mass $M$ and its acceleration $a_{\theta}$ in the $\hat \theta$ direction $$(3):= F=Ma_{\theta}$$
The acceleration of the particle in the $\hat \theta$ direction can be found by taking two time derivations of its position vector $\mathbf{r}$, and then taking the dot product with the unit vector $\hat \theta$. Doing that we get $$(4):= a_{\theta}=\ddot{\vec{r}} \cdot \hat \theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} $$
With $(4)$ we can write $(3)$ as $$F=M( r\ddot{\theta} + 2\dot{r}\dot{\theta})$$
and $(2)$ as $$(2)':=\tau= Mr( r\ddot{\theta} + 2\dot{r}\dot{\theta}) = Mr^2\ddot{\theta} + 2Mr\dot{r}\dot{\theta} $$
Now, let us finally compare $(1)$ and $(2)'$: $$ (1)= \tau = Mr^2\ddot{\theta} $$ $$ (2)'= \tau = Mr^2\ddot{\theta} + 2Mr\dot{r}\dot{\theta} $$
We see that $(1)$ and $(2)'$ can only be equal when $\dot r = \dot \theta = 0$, which is only true when the particle has no velocity.
What is going on here? Is $I_O = d^2M$ only valid when the particle has no velocity? What about a real rigid body of multiple particles, can the parallel axis theorem not be applied when the velocity is non-zero?