So lets say you have a vector field $X = (2x+y) \partial x + x \partial y$ and a diffeomorphism $f(x,y) = (x-2y,2x+y)$, then I need to know how to find the f-related vector fields of X, lets say $\overline{X}$. That is, those that meet:
$\overline{X} \circ f = \partial f \circ X \Rightarrow \overline{X} = \partial f \circ X \circ f^{-1}$
Since $f^{-1} = \left(\frac{x+2y}{5},\frac{y-2x}{5}\right)$
$X \circ f^{-1} = \left[2(x \circ f^{-1})+(y \circ f^{-1})\right] \partial (x \circ f^{-1}) + \left[x \circ f^{-1}\right] \partial (y \circ f^{-1}) = \left[2(\frac{x+2y}{5})+(\frac{y-2x}{5})\right] \partial (\frac{x+2y}{5}) + \left[\frac{x+2y}{5}\right] \partial (\frac{y-2x}{5}) = y (\frac{1}{5}\partial x +\frac{2}{5}\partial y) + \left[\frac{x+2y}{5}\right] (-\frac{2}{5}\partial x +\frac{1}{5}\partial y) = \frac{y-2x}{25}\partial x + \frac{12y+x}{25}\partial y$
And since $\partial f = (\partial x -2 \partial y , 2 \partial x +\partial y)$
$\overline{X} = \partial f \circ X \circ f^{-1} = (\partial x -2 \partial y , 2 \partial x +\partial y) \circ \left(\frac{y-2x}{25}\partial x + \frac{12y+x}{25}\partial y\right)$
And the above composition would give a two-component object, which differs from what I was trying to get, a vector field.
Please let me know how to approach this.
Edit1: So I've seen that I had two main mistakes in what I presented above.
The first mistake is that is not $\partial f$ but $d f$, which corresponds to the jacobian matrix.
The other mistake was related to $(x \circ f^{-1})$ and $(y \circ f^{-1})$
So the correct way to solve the problem would be:
$\overline{X} \circ f = d f \circ X \Rightarrow \overline{X} = d f \circ X \circ f^{-1}$
Since $f^{-1} = \left(\frac{x+2y}{5},\frac{y-2x}{5}\right)$
$X \circ f^{-1} = \left[2(x \circ f^{-1})+(y \circ f^{-1})\right] \partial x + \left[x \circ f^{-1}\right] \partial y = \left[2(\frac{x+2y}{5})+(\frac{y-2x}{5})\right] \partial x + \left[\frac{x+2y}{5}\right] \partial y = y \partial x + \frac{x+2y}{5}\partial y $
And the above result can be seen as the column vector $\begin{pmatrix}y \\ \frac{x+2y}{5} \end{pmatrix}$
And since $d f =\begin{pmatrix}\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y}\\\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y}\end{pmatrix} = \begin{pmatrix}1 & -2\\2 & 1\end{pmatrix}$
$\overline{X} = d f \circ X \circ f^{-1} = \begin{pmatrix}1 & -2\\2 & 1\end{pmatrix}\begin{pmatrix}y \\ \frac{x+2y}{5} \end{pmatrix} = \begin{pmatrix}\frac{y-2x}{5} \\ \frac{x+12y}{5} \end{pmatrix} = \frac{y-2x}{5}\partial x + \frac{x+12y}{5} \partial y$
I'll use the notation of An Introduction to Manifolds (second edition) written by Loring W.Tu (in particular, chapter one), so that $$X = (2x+y) \frac{\partial}{\partial x} + x \frac{\partial}{\partial y}$$
A vector field $\overline{X}$ is $f$-related to $X$ if and only if $f_{*}X=\overline{X}$. More precisely, $f_{*, ~p}(X_p) = \overline{X}_{f(p)}$ for every point $p$ in the domain of $X$. In our case, $\overline{X}_q=f_{*, ~p}(X_{p})$ where $p=f^{-1}(q)$.
Note that \begin{align} \ f_* = \begin{bmatrix} 1 & -2 \\ 2 & 1\end{bmatrix} \end{align}
in a sense that $$f_{*} \left( \frac{\partial}{\partial x}\bigg|_{p} \right) = \frac{\partial}{\partial x}\bigg|_{q} + 2\frac{\partial}{\partial y}\bigg|_{q} $$ $$f_{*} \left( \frac{\partial}{\partial y}\bigg|_{p} \right) = -2\frac{\partial}{\partial x}\bigg|_{q} +\frac{\partial}{\partial y}\bigg|_{q} $$
Write $$\overline{X}=f(x,y)\frac{\partial}{\partial x}+g(x,y)\frac{\partial}{\partial y} $$
Put $q=(a,b)$. Then $$\overline{X}_{q}=f(a,b)\frac{\partial}{\partial x}\bigg|_{q} +g(a,b)\frac{\partial}{\partial y} \bigg|_{q}$$
On the other hand, $p=f^{-1}(q)=\frac{1}{5}(a+2b, b-2a )$.
Thus \begin{align} X_p &= \left( \frac{ 2a+4b+b-2a}{5} \right) \frac{\partial}{\partial x}\bigg|_{p} + \left( \frac { a+2b }{5} \right)\frac{\partial}{\partial y}\bigg|_{p} \\ &=b \frac{\partial}{\partial x}\bigg|_{p} + \left( \frac { a+2b }{5} \right)\frac{\partial}{\partial y}\bigg|_{p}\end{align}
Finally, \begin{align} f_{*, ~p}(X_p) &=b \left(\frac{\partial}{\partial x}\bigg|_{q} + 2\frac{\partial}{\partial y}\bigg|_{q} \right)+ \left( \frac { a+2b }{5} \right) \left( -2\frac{\partial}{\partial x}\bigg|_{q} +\frac{\partial}{\partial y}\bigg|_{q}\right) \\ &= \left( \frac { -2a+b }{5} \right) \frac{\partial}{\partial x}\bigg|_{q} + \left( \frac { a+12b }{5} \right) \frac{\partial}{\partial y}\bigg|_{q} \end{align}
Therefore, $f(a,b) = \frac{1}{5}(-2a+b) $ and $g(a, b)= \frac{1}{5}(a+12b)$.
In short, $$\overline{X}= \left( \frac { -2x+y }{5} \right)\frac{\partial}{\partial x} +\left( \frac { x+12y }{5} \right)\frac{\partial}{\partial y}$$