According to tensor notation, to transform a vector from coordinates $x^\mu$ to $x^{\mu'}$, one applies the rule $$ V^{\mu'} = \frac{\partial x^{\mu'}}{\partial x^\mu} V^\mu $$
However, for example in polar coordinates, $$ \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2+y^2}} \;\; ; \;\; \frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2+y^2}} $$ $$ \frac{\partial \theta}{\partial x} = \frac{-y}{\sqrt{x^2+y^2}} \;\; ; \;\; \frac{\partial \theta}{\partial y} = \frac{x}{\sqrt{x^2+y^2}} $$ which yields, for the vector $V^\mu = (x,y)$ in cartesian coordinates, $$ V^\theta = x\frac{\partial \theta}{\partial x}+y\frac{\partial \theta}{\partial y}=0 $$
However, one would expect the vector described to be $V^{\mu'} = (r,\theta)$, ie $V^\theta=\theta$.
I understand that the above transformation is a matrix transformation, and thus holds for rotations, scalings etc. However, it doesn't work under a change to polar coordinates, as shown.
What is the failing assumption? The transformation law? (If so, when does it apply?) The assumption that $(x,y)$ is a vector? (If so, how can we define things like velocity as the derivative of position?)
Your expectation is incorrect. The vector $x\,\partial_x + y\,\partial_y$ points radially outward from the origin and has length equal to $\sqrt{x^2+y^2}=r$. In polar coordinates therefore, we should expect that the vector is $r\,\partial_r$. And it is. You've already shown that the $\theta$ component is zero. We can also compute the $r$ component in the same way: $$V^r = x\frac{\partial r}{\partial x} + y\frac{\partial r}{\partial y} = \frac{x^2+y^2}{\sqrt{x^2+y^2}} = \sqrt{x^2+y^2} = r$$ which agrees with the expectation.