Differential forms should be invariant under coordinate transformations I am wondering why, if we transform the following differential form, it does not seem to be invariant under the coordinate transformation. The $1$-form on $\mathbb R^2$ is
$$
 \omega = \sqrt{x^2 + y^2} dx + 0\cdot dy
$$
and as the coefficient function for $dx$ is radially symmetric, it should be easier to express this form in polar coordinates. Then with
$$
 dx = \cos\theta dr - r\sin\theta d\theta \quad
 dy = \sin\theta dr + r\sin\theta d\theta 
$$
and $f(x(r,\theta), y(r,\theta)) = r$ for $f(x,y) = \sqrt{x^2+y^2}$ we have
$$
 \omega = r(\cos\theta dr - r\sin\theta d\theta)
        = r\cos\theta dr - r^2\sin\theta d\theta.
$$
So, now it might be reasonable to expect that the value of this differential form is not changed by the coordinate transformation. So consider the point $(1,1)$ in cartesian coordinates, and the tangent vector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ to that point, then evaluating $\omega$ in cartesian coordinates we have
$$
 \omega(1,1)\left( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right)
  = \sqrt{2} \cdot dx\begin{pmatrix} 1 \\ 0 \end{pmatrix}
  = \sqrt{2}
$$
as $dx\begin{pmatrix} x \\ y \end{pmatrix} = x$. Now evaluating the form in polar coordinates, the same point is $(\sqrt 2, \pi/4)$ expressed in polar coordinates, so evaluating $\omega$ in polar coordinates we get
\begin{align*}
 \omega(\sqrt 2, \pi/4)\left( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right)
 & = \sqrt 2 \cos(\pi/4) dr\begin{pmatrix} 1 \\ 0 \end{pmatrix}
 - 2\sin(\pi/4) d\theta\begin{pmatrix} 1 \\ 0 \end{pmatrix} \\
 & = 1\cdot dr\begin{pmatrix} 1 \\ 0 \end{pmatrix} - \sqrt 2 \cdot d\theta\begin{pmatrix} 1 \\ 0 \end{pmatrix} \\
 & = 1 \cdot 0 - \sqrt 2 \cdot 0 \\
 & = 1
\end{align*}
which is different (again using that $dr, d\theta$ maps on the first and second component). So maybe the tangent vector has to be transformed too, so
$$
 \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \cos\theta \\ \sin\theta\end{pmatrix}
 = \frac{1}{2}\begin{pmatrix}\sqrt 2 \\ \sqrt 2 \end{pmatrix}.
$$
So now doing the second computation again:
\begin{align*}
 \omega(\sqrt 2, \pi/4)\left( \frac{1}{2} \begin{pmatrix} \sqrt 2 \\ \sqrt 2 \end{pmatrix} \right) 
 & = \sqrt 2 \cos(\pi/4) dr \left( \frac{1}{2} \begin{pmatrix} \sqrt 2 \\ \sqrt 2 \end{pmatrix} \right) - 2\sin(\pi/4) d\theta \left( \frac{1}{2} \begin{pmatrix} \sqrt 2 \\ \sqrt 2 \end{pmatrix} \right) \\
 & = \frac{1}{2}\sqrt 2 - \sqrt 2 \frac{1}{2}\sqrt 2 \\
 & = \frac{1}{2}\sqrt 2 - 1
\end{align*} 
which is again different.
So whats wrong here? Have I done any transformations wrong? The form should yield the same value for the same point and tangent vector, regardless of the coordinate system in which it is written, or not?

The problem with your computation is that the polar coordinates of the basis tangent vector $\partial_x$ at $(1,1)$ are $(\sqrt 2 /2,-1/2)$:
$$v = \partial_x\rvert_{(1,1)} = \frac{\partial r}{\partial x}\rvert_{(1,1)}\partial_r\rvert_{(1,1)} + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}\partial_\theta\rvert_{(1,1)} = \frac{\sqrt 2}{2}\partial_r\rvert_{(1,1)} - \frac{1}{2}\partial_\theta\rvert_{(1,1)}.$$
Therefore we find $\omega_{(1,1)}(v) = dr_{(1,1)}(v) - \sqrt 2 d\theta_{(1,1)}(v) = \frac{\sqrt2}{2} + \frac{\sqrt 2}{2} = \sqrt 2$.

Viewing $\mathbb{R}^2 - 0$ as a manifold, you have a polar coordinate chart $\phi = (r,\theta)$ which maps some neighborhood of $(1,1)$ onto a neighborhood of $(\sqrt 2,\pi/4)$. The basis vectors $\partial_r\rvert_{(1,1)}$ and $\partial_\theta\rvert_{(1,1)}$ of the tangent space at $(1,1)$ in terms of this coordinate chart are defined as pushforwards of the standard basis vectors of the tangent space at $(\sqrt 2,\pi/4)$:
$$\partial_r\rvert_{(1,1)} = (\phi^{-1})_*(\partial_x\rvert_{(\sqrt 2,\pi/4)}), \quad \partial_\theta\rvert_{(1,1)} = (\phi^{-1})_*(\partial_y\rvert_{(\sqrt 2,\pi/4)}).$$
By definition of the pushforward, we have $$\phi_*(\partial_x\rvert_{(1,1)}) = \frac{\partial r}{\partial x}\rvert_{(1,1)}\partial_x\rvert_{(\sqrt 2, \pi/4)} + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}\partial_y\rvert_{(\sqrt 2, \pi/4)}$$
and so
$$\partial_x\rvert_{(1,1)} = \frac{\partial r}{\partial x}\rvert_{(1,1)}(\phi^{-1})_*(\partial_x\rvert_{(\sqrt 2,\pi/4)}) + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}(\phi^{-1})_*(\partial_y\rvert_{(\sqrt 2,\pi/4)}) = \frac{\partial r}{\partial x}\rvert_{(1,1)}\partial_r\rvert_{(1,1)} + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}\partial_\theta\rvert_{(1,1)}.$$
