In general, the coordinates of a vector are defined as the projections of it onto the coordinate axis. Moreover, in a polar coordinate system, the basis vectors $\hat e_\phi$, $\hat e_r$ depend on the location.
For concreteness, let's assume that we want to describe an object that starts at $\vec a=(1,1)_{x,y}=(\sqrt{2},45^\circ)_{r,\phi}$ and ends at $\vec a=(1,-1)_{x,y}=(\sqrt{2},-45^\circ)_{r,\phi}$.
How can we understand the coordinates in the polar coordinate system in terms of projections onto the coordiante axis, $\hat e_\phi$, $\hat e_r$? In particular, which basis vectors do we use? (This is nontrivial since the basis vectors $\hat e_\phi$, $\hat e_r$ depend on the location. Therefore, if we use the basis vectors at the location our vectors point to, we always find zero for the projection onto $\hat e_\phi$ which is obviously wrong.)
The angle is usually called $\theta$ in $2$ dimensions.
The result $\vec{r}=r\vec{e}_r$ isn't a bug, it's a feature. Since $\vec{e}_r=\cos\theta\vec{e}_i+\sin\theta\vec{e}_j$ and $\vec{e}_\theta=-\sin\theta\vec{e}_i+\cos\theta\vec{e}_j$, $\frac{d}{d\theta}\vec{e}_r=\vec{e}_\theta$ and $\frac{d}{d\theta}\vec{e}_\theta=-\vec{e}_r$, so $\frac{d}{d\theta}\vec{r}=\frac{dr}{d\theta}\vec{e}_r+r\vec{e}_\theta$.
If $\theta$ is time-dependent, $\dot{\vec{r}}=\dot{\theta}\frac{d}{d\theta}\vec{r}$ by the chain rule. This is useful in kinematics, where we can also obtain $\ddot{\vec{r}}=\left(\ddot{r}-r\dot{\theta}^2\right)\vec{e}_r+\frac1r\frac{d}{dt}\left(r^2\dot{\theta}\right)\vec{e}_\theta$. If we move into three dimensions viz. the right-hand rule $\vec{e}_z:=\vec{e}_r\times\vec{e}_\theta$, $\vec{r}\times\dot{\vec{r}}=r^2\dot{\theta}\vec{e}_z$. Thus angular momentum is conserved if the acceleration is radial, which is very useful.
Edit: I'll discuss how we get basis vectors for general coordinates.
Let $x^i$ denote a Cartesian coordinate system and $q^i$ an alternative coordinate system, and define the covariant basis $\vec{b}_i:=\frac{\partial\vec{r}}{\partial q^i}$ and contravariant basis $\vec{b}^j:=\vec{\nabla}q^j$ so$$\vec{b}_i\cdot\vec{b}^j=\frac{\partial x^k}{\partial q^i}\frac{\partial q^j}{\partial x^k}=\frac{\partial q^j}{\partial q^i}=\delta_i^j.$$In the polar example$$q^1=r,\,q^2=\theta,\,\vec{b}_1=\vec{e}_r,\,\vec{b}_2=r\vec{e}_\theta,\,\vec{b}^1=\vec{e}_r,\,\vec{b}^2=\frac1r\vec{e}_\theta.$$