Let $T$ rotate every point through the same angle $\phi$ about the origin, $i.e.$ $T: (r, \theta) \to (r, \theta + \phi)$ where $\phi$ is given. If in addition that $T(O) = O,$ namely, if $T$ maps the zero element to itself, must $T$ be linear?
I think not so; for $T(ar + b\psi_{1}, a\theta + b\psi_{2}) = (ar + b\psi_{1}, a\theta + b\psi_{2} + \phi) = a(r, \theta) + b(\psi_{1}, \psi_{2}) + (0, \phi).$ Is this true?
The issue with your work is that $$ (ar+b\psi_1,a\theta+b\psi_2+\phi)\neq a(r,\theta)+b(\psi_1,\psi_2)+(0,\phi). $$ The second vector doesn't have the proper form of $(\text{radius},\text{angle})$. In general, adding polar coordinates isn't easy without converting to rectangular coordinates first.
You need to answer two questions. If you rotate a vector and then increase its magnitude, is that the same as increasing its magnitude and then rotating it? And, if you add two vectors and then rotate, is that the same as rotating those vectors and then adding them?
It is probably best to consider rectangular coordinates, and find a matrix that represents rotation by $\theta$. If you can find such a matrix, then the transformation will be linear.