A vector $\vec{\omega}$ in General Relativity is expressed in the language of differential geometry as $$\omega=\omega_r\frac{\partial}{\partial r}+\omega_\theta\frac{\partial}{\partial\theta}$$. However, I want to express this in the usual vector notation as $$\vec{\omega}=\omega(r)\hat{r}+\omega(\theta)\hat{\theta}$$ so as to obtain the magnitude of $\vec{\omega}$ as $\omega=\sqrt{\omega^2(r)+\omega^2(\theta)}$.
Could someone please suggest how to do this?
The tangent space at a point $p$ in a manifold consists of the Liebnizian operators (i.e., differential operators) on functions at $p$. By means of a coordinate chart $(x_i)_i$ about $p$, we see that the operators $\frac{\partial}{\partial x_i}$ are such Liebnizian operators, and in fact form a basis for the tangent space.
Now on $\Bbb R^n, \frac{\partial}{\partial x_i}$ is easily seen to be the directional derivative in the direction $\hat x_i$ and for a general vector $\vec v$ with $$\vec v = \sum_i v_i\hat x_i$$ we find that the directional derivative with respect to $\vec v$ is $$\frac{\partial}{\partial \vec v} = \sum_i v_i\frac{\partial}{\partial x_i}$$ Thus we can associate $\vec v$ with its directional derivative $\frac{\partial}{\partial \vec v}$. This same identification is also made for manifolds.
In short, $$\frac{\partial}{\partial r} \equiv \hat r\\\frac{\partial}{\partial \theta} \equiv r\hat \theta$$ Per the comment by peek-a-boo, $\frac{\partial}{\partial \theta}$ is not a unit vector. Its length is $r$. Thus the unit vector $\hat \theta$ is multiplied by $r$.