Take some general vector $\mathbf{u}$ which can be expressed in terms of is components and basis as $\mathbf{u} = u^{a} e_a$ (summation notation).
Now, if we transform to some different, primed basis, then the contravariant vector components transform as,
$$u^{a'} = \frac{\partial x^{a'}}{\partial x^b} u^b$$
In this way we can transform from e.g. Cartesian to polar coordinates. That is all fine
Now my confusion comes where if we have some vector defined at the origin. For example we have a velocity vector $\mathbf{u}$ with components $(u^x, u^y)$ which 'exists' at $x=0, y=0$.
If we then attempt to go to polar coordinates then
$$u^r = \frac{\partial r}{\partial x} u^x + \frac{\partial r}{\partial y} u^y $$
$$ = \frac{x}{r} u^x + \frac{y}{r} u^y$$
$$ = 0$$
but then , intuitively, I feel like the vector $\mathbf{u}$ should have some component in the $r$-direction, since $u^x, u^y$ are non zero? What gives?
Assume ${\bf{u}} = u_x {\bf{e_x}} + u_y {\bf{e_y}}$ with the cartesian basis vectors ${\bf{e_x}}$ and ${\bf{e_y}}$ with $u_x\neq0$ and $u_y\neq 0$. Then ${\bf{u}} = u_r {\bf{e_r}} + u_\vartheta {\bf{e_\vartheta}}$ in polar coordinates.
Because ${\bf{e_r}}=\cos(\vartheta)\bf{e_x}+\sin(\vartheta){\bf{e_y}}$ and ${\bf{e_\theta}}=-\sin(\vartheta)\bf{e_x}+\cos(\vartheta)\bf{e_y}$ we can substitute in the polar representation and we get
$$\begin{align*}{\bf{u}} &=u_r {\bf{e_r}} + u_\vartheta {\bf{e_\vartheta}} \\&= u_r (\cos(\vartheta){\bf{e_x}}+\sin(\vartheta){\bf{e_y}}) + u_\vartheta (-\sin(\vartheta)\bf{e_x}+\cos(\vartheta)\bf{e_y}) \\&= (u_r \cos(\vartheta)-u_\vartheta \sin(\vartheta)){\bf{e_x}}+(u_r \sin(\vartheta) + u_\vartheta \cos(\vartheta)){\bf{e_y}}\\&=u_x {\bf{e_x}} + u_y {\bf{e_y}} \end{align*}$$
From the last two equations we can read the (contravariant) coordinate trasformation of the vector components:
$$\begin{align*}u_x &=u_r \cos(\vartheta)-u_\vartheta \sin(\vartheta) \\u_y &= u_r \sin(\vartheta) + u_\vartheta \cos(\vartheta) \end{align*}$$
or
$$\begin{align*}u_r &=u_x \cos(\vartheta)+u_y \sin(\vartheta) \\u_\vartheta &= -u_x\sin(\vartheta) + u_y \cos(\vartheta) \end{align*}$$
EDIT
if $\vartheta = \arctan(u_y/u_x)$, then $\cos(\vartheta) = \frac{1}{\sqrt{1+\left(u_y/u_x\right)^2}}=\frac{u_x}{\sqrt{u_x^2+u_y^2}}$ and $\sin(\vartheta) = \frac{u_y/u_x}{\sqrt{1+\left(u_y/u_x\right)^2}} = \frac{u_y}{\sqrt{u_x^2+u_y^2}}$ then
$$\begin{align*}u_r &= u_x \cos(\vartheta)+u_y \sin(\vartheta) \\&=u_x \frac{1}{\sqrt{1+\left(u_y/u_x\right)^2}}+u_y \frac{u_y/u_x}{\sqrt{1+\left(u_y/u_x\right)^2}} \\&=\sqrt{u_x^2+u_y^2} \end{align*}$$