Aren't All Bases Othonormal?

You can't just switch bases and expect the dot product to compute the same thing, as your example shows. The coordinate representation of the inner product has to change away from the dot product to something else.

The dot product's Gram matrix reads $\begin{bmatrix}1&0\\0&1\end{bmatrix}$ but after the change of basis which converts from these coordinates to coordinates of the second basis you gave, the new Gram matrix is

$\begin{bmatrix}4&8\\8&20\end{bmatrix}$, thus the inner product, computed both with the old and new coordinates matches:

$\begin{bmatrix}2&0\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}4\\2\end{bmatrix}=\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}4&8\\8&20\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=8$.

The mistake you were making was thinking that the Gram matrix stayed the same between the two expressions, but it does change.

And what about the original basis?

Well $[1,0]$ was sent to $[1/2, 0]$ and $[0,1]$ was sent to $[-1, 1/2]$, and

$\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}1/2&0\end{bmatrix}\begin{bmatrix}4&8\\8&20\end{bmatrix}\begin{bmatrix}-1\\1/2\end{bmatrix}=0$

$\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}1/2&0\end{bmatrix}\begin{bmatrix}4&8\\8&20\end{bmatrix}\begin{bmatrix}1/2\\0\end{bmatrix}=1$

$\begin{bmatrix}0&1\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}-1&1/2\end{bmatrix}\begin{bmatrix}4&8\\8&20\end{bmatrix}\begin{bmatrix}-1\\1/2\end{bmatrix}=1$

so it is still orthonormal.

Remember: Inner products have an existence outside of coordinates, and the dot product depends on a basis. We pick a basis and switch to coordinates to do computation sometimes, but choice of basis is immaterial, and won't affect the end result that the inner product computes. So, changing the basis out from under the inner product changes the way the quantity is computed.

The same goes for linear transformations and the matrices we choose to represent them.

To give an example of a similar mistake you could try with transformations:

"$\begin{bmatrix}1\\1\end{bmatrix}$ is in the kernel of the linear transformation $\begin{bmatrix}1&-1\\1&-1\end{bmatrix}$, but when I use $v=\begin{bmatrix}1\\1\end{bmatrix}$ in a basis so that $v=\begin{bmatrix}1\\0\end{bmatrix}$ it is not in the kernel anymore... why?"

The answer is, of course, that the matrix has to change when the basis changes too.