If the radius in polar coordinates is $r(\theta)$, then the position vector is $\mathbf{r}=(x,y)$, where
\begin{eqnarray}
x &=& r(\theta) \cos \theta \\
y &=& r(\theta) \sin \theta
\end{eqnarray}
I am assuming your problem is to consider $\theta=\omega t$ (if this is true, please edit your question to clarify). These sorts of calculations are always the most straightforward in Cartesian coordinates. Then we can take derivatives to yield
\begin{eqnarray}
x &=& r_0 \left( 1-\cos\omega t \right) \cos\omega t \\
y &=& r_0 \left( 1-\cos\omega t \right) \sin\omega t
\end{eqnarray}
\begin{eqnarray}
\dot{x} &=& r_0 \omega \left( \sin2\omega t -\sin\omega t \right) \\
\dot{y} &=& r_0 \omega \left( -\cos2\omega t +\cos\omega t \right)
\end{eqnarray}
\begin{eqnarray}
\ddot{z} &=& r_0 \omega^2 \left(2\cos2\omega t-\cos\omega t \right) \\
\ddot{y} &=& r_0 \omega^2 \left(2\sin2\omega t -\sin\omega t \right)
\end{eqnarray}
Now, if you want to write these quantities as vectors, one can do so in terms of the Cartesian unit vectors $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$. For example, the velocity is
$$\dot{\mathbf{r}} = r_0 \omega \left( \sin2\omega t -\sin\omega t \right)\hat{\mathbf{x}} + r_0 \omega \left( -\cos2\omega t +\cos\omega t \right)\hat{\mathbf{y}}$$
It is also possible to do the entire calculation with the polar unit vectors $\hat{\mathbf{r}}$ and $\hat{\mathbf{\theta}}$, where
\begin{eqnarray}
\hat{\mathbf{r}} &=& \cos\theta \, \hat{\mathbf{x}} +\sin\theta \,\hat{\mathbf{y}} \\
\hat{\mathbf{\theta}} &=& \cos\theta \, \hat{\mathbf{y}} -\sin\theta \,\hat{\mathbf{x}}
\end{eqnarray}
\begin{eqnarray}
\hat{\mathbf{x}} &=& \cos\theta \, \hat{\mathbf{r}} -\sin\theta \,\hat{\mathbf{\theta}} \\
\hat{\mathbf{y}} &=& \sin\theta \, \hat{\mathbf{r}} +\cos\theta \,\hat{\mathbf{\theta}}
\end{eqnarray}
For this problem, since $\theta = \omega t$, the derivatives of the polar unit vectors are simple:
\begin{eqnarray}
\dot{\hat{\mathbf{r}}} &=& \omega \hat{\mathbf{\theta}} \\
\dot{\hat{\mathbf{\theta}}} &=& -\omega \hat{\mathbf{r}}
\end{eqnarray}
And then, because of the simplicity of the derivatives of the unit vectors, we can show
\begin{eqnarray}
\mathbf{r}/r_0 &=& \left(1-\cos\omega t \right)\hat{\mathbf{r}} \\
\dot{\mathbf{r}}/(r_0\omega) &=& \sin\omega t \, \hat{\mathbf{r}}+(1-\cos\omega t ) \hat{\mathbf{\theta}} \\
\ddot{\mathbf{r}}/(r_0\omega^2) &=& (2\cos\omega t-1) \, \hat{\mathbf{r}}+2\sin\omega t \, \hat{\mathbf{\theta}}
\end{eqnarray}
