A lot of confusion seems to come from mixing up vectors (and tensors) with vector (or tensor) fields. In a lot of literature the word field is omitted for brevity.
In two dimensions the coordinate transformation between cartesian and polar coordinates is
$$
\left(\begin{matrix}x\\y\end{matrix}\right)\mapsto\left(\begin{matrix}r\\\varphi\end{matrix}\right)=\left(\begin{matrix}\sqrt{x^2+y^2}\\\arctan(\frac{y}{x})\end{matrix}\right)\,.
$$
The back transformation is
$$
\left(\begin{matrix}r\\\varphi\end{matrix}\right)\mapsto\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}r\cos\varphi\\r\sin\varphi\end{matrix}\right)\,.
$$
These maps are one to one when restricted to $M=\mathbb R^2\setminus\{0\}\,.$
So far the above maps only express points on the manifold in different languages. Recall that a tensor field on the manifold is a map from $M$ into the tensor product of a selection of (co-)tangential spaces. For example, a contravariant vector field is
$$\tag{1}
V=v^x\partial_x+v^y\partial_y
$$
where $v^x,v^y$ are functions on $M$ and $\partial_x,\partial_y$ are a basis of $T_pM\simeq\mathbb R^2\,.$ The $x,y$-notation indicates that $V$ is expressed in cartesian coordinates. To express $V$ in polar coordinates we use the chain rule:
\begin{align}
\partial_x&=\frac{\partial r}{\partial x}\partial_r+\frac{\partial \varphi}{\partial x}\partial_\varphi=\frac{x}{\sqrt{x^2+y^2}}\partial_r-\frac{y}{x^2+y^2}\partial_\varphi=\cos\varphi\,\partial_r-\frac{\sin\varphi}{r}\,\partial_\varphi\,,\\
\partial_y&=\frac{\partial r}{\partial y}\partial_r+\frac{\partial \varphi}{\partial y}\partial_\varphi=\frac{y}{\sqrt{x^2+y^2}}\partial_r+\frac{x}{x^2+y^2}\partial_\varphi=\sin\varphi\,\partial_r+\frac{\cos\varphi}{r}\,\partial_\varphi\,.
\end{align}
Plugging this into (1) gives us the vector field in polar coordinates:
$$\tag{2}
V=\underbrace{(v^x\cos\varphi\,+v^y\sin\varphi)}_{\displaystyle =:v^r}\,\partial_r+\underbrace{\frac{-v^x\sin\varphi+v^y\cos\varphi}{r}}_{\displaystyle=:v^\varphi}\,\partial_\varphi\,.
$$
To shed some light on the concept of $\partial_x,\partial_y$ being vector fields recall that, for a real valued function $f$ on $M$ and a curve $\gamma$ passing through $p=\gamma(0)\,,$ the chain rule gives for the derivative of $f$ along the tangent of the curve
\begin{align}
\frac{d}{dt}\Bigg|_{t=0}f(\gamma(t))&=\partial_xf(\gamma(0))\,\gamma_1'(0)+\partial_yf(\gamma(0))\,\gamma_2'(0)\\
&=\partial_xf(p)\,\gamma_1'(0)+\partial_yf(p)\,\gamma_2'(0)\\
%&=\langle\nabla f(p),\gamma'(0)\rangle\,.
\end{align}
If we remove the test function $f$ from this relationship we get the RHS as
$$
\gamma_1'(0)\,\partial_x|_p+\gamma_2'(0)\,\partial_y|_p\,.
$$
This shows that $\gamma_1'(0),\gamma_2'(0)$ are the components of a general tangential vector in $T_pM$, and $\partial_x|_p,\partial_y|_p$ are the basis vectors.
See also Tangent Space.
