The vector function and its derivatives for a non uniform circular motion is:
$$ \vec s(t) = r \cdot \begin{bmatrix} \cos(\omega t) \\ \sin(\omega t) \end{bmatrix}, \qquad \omega = \omega(t) $$
$$ \vec v(t) = r \cdot \left({d\omega \over dt} t + \omega\right) \begin{bmatrix} -\sin(\omega t) \\ \cos(\omega t) \end{bmatrix} $$
$$ \vec a(t) = -r \cdot \begin{bmatrix} \left({d^2\omega \over dt^2} t + 2{d\omega \over dt}\right) \sin(\omega t) + \left({d\omega \over dt} t + \omega\right)^2 \cos(\omega t) \\ -\left({d^2\omega \over dt^2} t + 2{d\omega \over dt}\right) \cos(\omega t) + \left({d\omega \over dt} t + \omega\right)^2 \sin(\omega t) \end{bmatrix} $$
Something weird about $\vec v$
As $t$ approaches $∞$ and ${d\omega \over dt} \ne 0$, $|\vec v|$ also approaches $∞$.
That's no reason for speed to approach $∞$.
Consider this $$ \omega(t) = {1 \over r}{d\ell(t) \over dt}, \qquad \mathrm{where \ \ell(t) \ is \ the \ circle's \ traveled \ arc \ length \ after \ time \ t } $$
$${d\ell(t) \over dt} = |\vec v(t)| \quad \mathrm{gives} \quad |\vec v(t)| = r \omega(t) $$
But $ |\vec v(t)| = r \cdot \left({d\omega \over dt} t + \omega\right) $ !!!
Your questions seem to imply that you're under the impression that your quantity $\omega$ is the instantaneous angular velocity. This is not the case, as you can see e.g. from the fact that, as you noted, $|\vec v(t)|=r\omega(t)$ doesn't hold, whereas this holds for the instantaneous angular velocity.
To describe a non-uniform circular motion, use
$$ \vec s(t) = r \cdot \begin{bmatrix} \cos(\phi (t)) \\ \sin(\phi(t)) \end{bmatrix}\;; $$
then
\begin{align} \vec v(t) &= r\dot\phi(t) \cdot \begin{bmatrix}-\sin(\phi (t)) \\ \cos(\phi(t)) \end{bmatrix}\\ &=r\omega(t)\cdot \begin{bmatrix}-\sin(\phi (t)) \\ \cos(\phi(t)) \end{bmatrix}\;. \end{align}