Expanding on Ted Shifrin's comment:
It is shown in this answer
that the relationship between the components of vector fields $V$ and $U$ in
Cartesian and polar coordinates are:
\begin{align}
V&=v^x\partial_x+v^y\partial_y\\
&=\underbrace{(v^x\cos\color{red}\varphi\,+v^y\sin\color{red}\varphi)}_{\displaystyle =:v^r}\,\partial_r+\underbrace{\frac{-v^x\sin\color{red}\varphi+v^y\cos\color{red}\varphi}{\color{red}r}}_{\displaystyle=:v^\varphi}\,\partial_\varphi\,,\\
U&=u^x\partial_x+u^y\partial_y\\
&=\underbrace{(u^x\cos\color{red}\varphi\,+u^y\sin\color{red}\varphi)}_{\displaystyle =:u^r}\,\partial_r+\underbrace{\frac{-u^x\sin\color{red}\varphi+u^y\cos\color{red}\varphi}{\color{red}r}}_{\displaystyle=:u^\varphi}\,\partial_\varphi\,.
\end{align}
$v^x,v^y,u^x,u^y$ are the Cartesian components.
$v^r,v^\varphi,u^r,u^\varphi$ are the polar components.
In both systems the upper indices are just labels.
In contrast, the highlighted $\color{red}{r,\varphi}$ above are the coordinates of the point $\color{red}p$ on the manifold $\mathbb R^2$ at which both vectors $V,U$ must reside in order to be additible.
Then,
\begin{align}
V+U&=(v^x+u^x)\,\partial_x+(v^y+u^y)\,\partial_x\,\\[2mm]
&=(v^r+u^r)\,\partial_r+(v^\varphi+u^\varphi)\,\partial_\varphi\,,\\[2mm]
v^r+u^r&=(v^x+u^x)\cos\color{red}\varphi+(v^y+u^y)\sin\color{red}\varphi\,,\\[2mm]
v^\varphi+u^\varphi&=\frac{-(v^x+u^x)\sin\color{red}\varphi+(v^y+u^y)\cos\color{red}\varphi}{\color{red}r}\,.
\end{align}
- Regardless in which coordinate system we are: addition of vectors at the same point $\color{red}p$ is equivalent to addition of components. Anything else would be a nightmare.
Since there seems a disagreement with the answer by TurlocTheRed I elaborate further.
- The origin does not have polar coordinates so we cannot consider vectors that reside at $\color{red}p=(0,0)\,.$ Instead I shall consider
$$
\color{red}p=(1,1)=(\sqrt{2}\,;\pi/4)
$$
using the convention that the comma separates cartesian coordinates, and the semicolon separates polar coordinates. The same convention shall apply to
vector components. This point has polar coordinates
\begin{align}
\color{red}r&=\sqrt{2},\quad\color{red}\varphi=\pi/4\,,&\cos\color{red}\varphi
&=\sin\color{red}\varphi=\frac{1}{\sqrt{2}}\,.
\end{align}
With the formulas above one can easily check that in both coordinates systems
components add when we add the vectors:
\begin{align}
V_\color{red}p&=(1,1)=(\sqrt{2}\,;0)\,,&U_\color{red}p&=(0,1)=\textstyle(\frac{1}{\sqrt{2}};\frac12)\,,&V_\color{red}p+U_\color{red}p&=(1,2)=\textstyle(\frac{3}{\sqrt{2}};\frac12)\,.
\end{align}
- The polar components of these vectors could be a bit puzzling at first but note that $V_\color{red}p$ points from $(1,1)$ to $(2,2)\,.$ This means we don't go in an angular direction and only in the radial direction. That's why the polar components of $V_\color{red}p$ are $(\sqrt{2}\,;0)\,.$
