Your assumption for the velocity vector to be contravariant is kind of wrong.
You can define any vector in the following manner,
$$\vec{V}=V^i\mathbf{e}_i=V_i\mathbf{e}^i $$
In case we want to assume contravariant components, we also need to define set of covariant basis. We can do that through the covariant transformation, which is,
$${V'}_i=\frac{\partial{x'}_j}{\partial x_i}V_j $$
In this case we shall get the two covariant basis as,
$$\mathbf{e}_r=\frac{\partial x}{\partial r}\mathbf{e}_x+\frac{\partial y}{\partial r}\mathbf{e}_y $$
$$\mathbf{e}_{\theta}=\frac{\partial x}{\partial \theta}\mathbf{e}_x+\frac{\partial y}{\partial\theta}\mathbf{e}_y $$
That would give us,
$$\mathbf{e}_r=\cos\theta\, \mathbf{e}_x+\sin\theta\, \mathbf{e}_y $$
$$\mathbf{e}_{\theta}=-r\sin\theta\, \mathbf{e}_x+r\cos\theta\, \mathbf{e}_y $$
Note that these aren't unit basis vector, which is the one generally introduced at the pre-tensor level. If you consider the inner product of each of the basis in the Cartesian coordinate then we shall get,
$$\mathbf{e}_r\cdot\mathbf{e}_r=1 $$
$$\mathbf{e}_{\theta}\cdot\mathbf{e}_{\theta}=r^2 $$
Which also means that the tangential basis, is not a unit basis. The position vector in this case will be,
$$\vec{r}=r\, \mathbf{e}_r $$
Which makes the velocity vector,
$$\vec{v}=\frac{d\vec{r}}{dt}=\dot{r}\, \mathbf{e}_r+r\left(\frac{\dot{\theta}}{r}\right) \, \mathbf{e}_{\theta}$$
$$\Rightarrow \vec{v}=\dot{r}\, \mathbf{e}_r+\dot{\theta} \, \mathbf{e}_{\theta}$$
Note that the dimension of this vector is that of velocity. The tangential basis has the dimension of length which makes the total expression having a dimension of velocity. And now you can use the metric tensor while taking the dot product to see the invariant scalar which will be,
$$\vec{v}\cdot\vec{v}=\eta_{\alpha\beta}v^{\alpha}v^{\beta}=\dot{r}^2+r^2\dot{\theta}^2 $$
When you normalize the covariant basis (in this case also called vector basis, while the contravariant basis will be called one form basis) you are basically changing the basis definition from a coordinate basis to a noncoordinate basis. Noncoordinate basis cannot be obtained from coordinate transformation, which is why it isn't used in most of the cases of tensor application.