Consider two bases $e_i$ and $f_i$ of $\mathbb{R}^2$ defined by:
$$\begin{aligned}
(f_1,f_2) &= (e_1,e_2)\cdot F\\
\begin{pmatrix}1&-1\\1&1\end{pmatrix} &= \begin{pmatrix}1&0\\0&1\end{pmatrix} \cdot \begin{pmatrix}1&-1\\1&1\end{pmatrix}
\end{aligned}$$
and the cooresponding change of coordinates ($B\equiv F^{-1}$):
$$\begin{aligned}
B\cdot \begin{pmatrix}x\\y\end{pmatrix} &= \begin{pmatrix}\tilde x\\ \tilde y \end{pmatrix}_f = \begin{pmatrix}1/2 x + 1/2 y \\ -1/2 x+1/2y\end{pmatrix}_f \\
F\cdot \begin{pmatrix}\tilde x\\ \tilde y \end{pmatrix}_f &= \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}\tilde x-\tilde y\\ \tilde x+\tilde y\end{pmatrix}
\end{aligned}$$
I would like to find the gradient of a function in both coordinate systems. Here, tilde symbols are associated to the $f$-basis. Consider
$$\begin{aligned}
V(x,y) &= x^2 +y \\
\tilde V(\tilde x,\tilde y) &= (\tilde x-\tilde y)^2 + (\tilde x + \tilde y)
\end{aligned}$$
The gradients are defined using the metric-tensor scalings, as
$$\begin{aligned}
\nabla &= \partial_x e_1 + \partial_y e_2\\
\tilde \nabla &= \frac{1}{\sqrt{f_1\cdot f_1}} \partial_{\tilde x} f_1 +\frac{1}{\sqrt{f_2\cdot f_2}} \partial_{\tilde y} f_2 = \frac{1}{\sqrt{2}} \partial_{\tilde x} f_1 +\frac{1}{\sqrt{2}} \partial_{\tilde y} f_2
\end{aligned}$$
Lets check the gradients at the point $(2,1)\equiv (1.5, -0.5)_f$ where the second coordinates are with respect to the $f$-basis:
$$\begin{aligned}
\nabla V \vert_{(2,1)} &= \begin{pmatrix}2x \\1 \end{pmatrix}= \begin{pmatrix}4\\ 1\end{pmatrix} \\
\tilde \nabla\tilde V \vert_{(1.5,-0.5)}&= \frac{1}{\sqrt{2}} [2(\tilde x-\tilde y)+1] f_1 + \frac{1}{\sqrt{2}} [-2(\tilde x-\tilde y)+1] f_2 \\
&= \frac{1}{\sqrt{2}} 5 \begin{pmatrix}1\\1 \end{pmatrix} + \frac{1}{\sqrt{2}} (-3) \begin{pmatrix}-1\\1 \end{pmatrix} \\
&= \frac{1}{\sqrt{2}} \begin{pmatrix}8\\2 \end{pmatrix}
\end{aligned}$$
Why do i get a different gradient vector as a result?
Best Answer
Your gradient formula is incorrect; it should be $1/f_1\cdot f_1$ and $1/f_2\cdot f_2$ without the square roots. Then you get $$ \frac12\begin{pmatrix}8\\2\end{pmatrix} = \begin{pmatrix}4\\1\end{pmatrix}. $$
(It is also bad practice to write $\partial_{\tilde x}f_1$ as you do since we can and often do consider gradients in variable coordinate systems, and in this case the basis vectors $f_i$ will vary from point to point. The notation $\partial_{\tilde x}f_1$ suggest then that you would differentiate $f_1$, but this is incorrect when calculating the gradient and so the notation $f_1\partial_{\tilde x}$ is more advisable.)
The general procedure for any basis is to find its reciprocal basis $f^i$ defined by $$ f^i\cdot f_j = \delta^i_j. $$ Then the gradient is $$ \nabla = \sum_if^i\partial_i $$ where $\partial_i$ is the partial derivative with respect to the coordinate corresponding to $f_i$.
In your case your $f_i$ are orthogonal, so they are almost there own reciprocal; we just need to get the scaling right, and it should be obvious that $$ \frac{f_i}{f_i\cdot f_i}\cdot f_i = 1 $$ so indeed $f^i = f_i/f_i\cdot f_i$.