So, I talked to my professor today, asking him for a way to compute flows of vectorfields. An example was given by $ v = x\partial_x + y\partial_y$ living in $\mathbb{R}^2$. Now I know that the way to go would be to solve the system of ODE's given by $x'(t) = x(t), y'(t)=y(t)$ given varying initial values.
However, drawing the vectorfield one sees how it acts on rays that turn radially, so it looks like an easy one-dimensional problem when only looking at the rays. And since it is all radial we went on to look at the problem in polar coordinates. Form here on I get a little lost:
Let $\Phi(r,\varphi) := (r\cos\varphi, r\sin\varphi)$. We next want to get a description of $v$ in these coordinates. As I see it we cannot compute $\Phi_*(v)$ simply because at first $v$ is given in the wrong coordinates. I thought what we want is somewhat $\Phi^{-1}_*(v)$ and then replacing $x=r\cos\varphi$ a.s.o.
What we did however was something along these lines: $$\begin{align}x\partial_x + y\partial_y &= r\cos(\varphi)\partial(r\cos\varphi) + r\sin\varphi \partial(r\cos\varphi)\\ &= r\cos(\varphi)\left(\frac{\partial\Phi_1(r,\varphi)}{\partial r}\partial_r + \frac{\partial\Phi_1(r,\varphi)}{\partial \varphi}\partial_\varphi\right) + r\sin\varphi \left(\frac{\partial\Phi_2(r,\varphi)}{\partial r}\partial_r + \frac{\partial\Phi_2(r,\varphi)}{\partial \varphi}\partial_\varphi\right)\\ &= r\partial_r \end{align}$$
And then we have found the "pushforward" of $v$ somewhat magically. Now I do not get why this works? Why is this exactly what we want? These seem like magical calculations that randomly give the right result to me. Maybe I remember his steps wrong, but in the end we never computed $\Phi^{-1}$ or the inverse of the jacobian and yet came to the right result.
(The next steps would be to get the flow in the coordinates for $\Phi$ and write it back in the coordinates of $(x,y)$ I guess).
I would be happy for clarifications on why this should work, where it comes from and also maybe a more formal proof on how to compute the flow of the given vector field when first changing coordinates and then getting back.
It looks like your professor is exploiting the standard metric and the musical isomorphisms to avoid calculating the inverse matrix, because pulling back differential forms $(x,y)$ to $(r,\theta)$ works the same direction as we want. We just need to be able to freely change vector fields to forms, and that is provided by the metric (= inner product, or if want indefinite signatures, = nondegenerate symmetric bilinear form) inducing a natural isomorphism between tangent and cotangent vectors $v\leftrightarrow v^\flat:=\langle v,-\rangle$: if the metric in $x^i$ coordinates is $\sum_i h^i(\mathrm{d}x^i)^2$ then $(\partial_{x_i})^\flat = h^i\,\mathrm{d}x^i$ and the other way is $(\mathrm{d}x^i)^\sharp=\frac1{h^i}\partial_{x_i}$.
Now what you have given looks like the usual calculation for $x\,\mathrm{d}x+y\,\mathrm{d}y=r\,\mathrm{d}r$. Since the standard metric $\mathrm{d}x^2+\mathrm{d}y^2=\mathrm{d}r^2+r^2\,\mathrm{d}\theta^2$, both $(x,y)$ and $(r,\theta)$ are orthogonal coordinates making the calculation easy. So "sharp"ing it gives $x\partial_x+y\partial_y=r\partial_r$ indeed.
If we try a similar approach on $x\partial_y-y\partial_x$, we have $$x\,\mathrm{d}y-y\,\mathrm{d}x=r^2\,\mathrm{d}\theta$$ so when you sharp it, you need to remember the coefficient $r^2$ in front of $\mathrm{d}\theta^2$ in $\mathrm{d}r^2+r^2\mathrm{d}\theta^2$ means $(r^2\,\mathrm{d}\theta)^\sharp=\partial_\theta$ so the full calculation goes $$ x\partial_y-y\partial_x=(x\,\mathrm{d}y-y\,\mathrm{d}x)^\sharp=(r^2\mathrm{d}\theta)^\sharp=\partial_\theta. $$