Inner product and orthogonality in non orthogonal basis

The key point to understand here is that you really are dealing with two $\mathbb R^2$ here, although it's not that obvious when using the standard basis.

The first $\mathbb R^2$ is your vector space. Let's write this vector space and everything in it in blue. This first $\color{blue}{\mathbb R^2}$ is equipped with a vector space structure and additionally with the dot product $\color{blue}{\mathbf x\cdot\mathbf y = x_1y_1+x_2y_2}$.

Now as soon as you choose a basis $\{\color{blue}{\mathbf b_1},\color{blue}{\mathbf b_2}\}\subset\color{blue}{\mathbb R^2}$, you can write every vector $\color{blue}{\mathbf x}\in\color{blue}{\mathbb R^2}$ in an unique way as $\color{blue}{\mathbf x}=\color{red}{\xi_1}\color{blue}{\mathbf b_1}+\color{red}{\xi_2}\color{blue}{\mathbf b_2}$. Note that $\color{red}{\xi_1}$ and $\color{red}{\xi_2}$ are not the components of the vector in $\color{blue}{\mathbb R^2}$, but are base dependent.

But of course you need always two of them, and when doing vector addition and multiplication with scalar, you'll find they behave exactly like the components of a vector should behave. Therefore it does make sense to consider them as part of a $\color{red}{\mathbb R^2}$ which however is a different $\mathbb R^2$ than the original $\color{blue}{\mathbb R^2}$ we started with. In particular, the coordinate $\color{red}{\mathbb R^2}$ is not pre-equipped with an inner product.

The basis then defines a linear map $\beta$ from the coordinate $\color{red}{\mathbb R^2}$ to the original $\color{blue}{\mathbb R^2}$ given by $$\beta(\color{red}{\boldsymbol\xi})=\color{red}{\xi_1}\color{blue}{\mathbf b_1} + \color{red}{\xi_2}\color{blue}{\mathbf b_2}.$$

Now remember that I said that the coordinate $\color{red}{\mathbb R^2}$ is not pre-equipped with an inner product. That doesn't mean we cannot give it one. but we want to do it in a way that the product is preserved by the map $\beta$, that is, you want to have $$\color{red}{\langle \boldsymbol\xi,\boldsymbol\eta\rangle} = \beta(\color{red}{\boldsymbol\xi})\color{blue}{\cdot}\beta(\color{red}{\boldsymbol\eta})$$ where $\color{red}{\langle \boldsymbol\xi,\boldsymbol\eta\rangle}$ denotes the inner product in the coordinate $\color{red}{\mathbb R^2}$. By inserting the explicit formula of $\beta$, one easily sees that $$\color{red}{\langle \boldsymbol\xi,\boldsymbol\eta\rangle} = \sum_{j,k=1}^2 (\color{blue}{\mathbf b_j\cdot\mathbf b_k})\color{red}{\xi_j\eta_k}.$$

Now quite obviously, if $\{\color{blue}{\mathbf b_1},\color{blue}{\mathbf b_2}\}$ is not an orthogonal basis, then $\color{red}{\langle \boldsymbol\xi,\boldsymbol\eta\rangle}\ne\color{red}{\xi_1\eta_1}+\color{red}{\xi_2\eta_2}$, and indeed, the inner product on the coordinate $\color{red}{\mathbb R^2}$ explicitly depends on the chosen basis $\{\color{blue}{\mathbf b_1},\color{blue}{\mathbf b_2}\}$. But that is not really surprising, because the vector in $\color{blue}{\mathbb R^2}$ those coordinates describe does depend on the basis chosen, and of course different vectors in general have different inner products.

Note that by definition of the inner product, with $\beta(\color{red}{\boldsymbol\xi})=\color{blue}{\mathbf x}$ and $\beta(\color{red}{\boldsymbol\eta})=\color{blue}{\mathbf y}$ it is of course still true that $$\color{red}{\langle \boldsymbol\xi,\boldsymbol\eta\rangle} = \color{blue}{\textbf x\cdot\textbf y} = \color{blue}{x_1y_1} + \color{blue}{x_2y_2}.$$ But in general, $\color{blue}{x_1y_1} + \color{blue}{x_2y_2} \ne \color{red}{\xi_1\eta_1}+\color{red}{\xi_2\eta_2}$.

However if you chose the standard basis $\color{blue}{\mathbf b_k}{\mathbf e_k}$ then you obviously have $\color{red}{\xi_k}=\color{blue}{x_k}$ and $\color{red}{\langle \boldsymbol\xi,\boldsymbol\eta\rangle} = \color{blue}{x_1y_1+x_2y_2} = \color{red}{\xi_1\eta_1}+\color{red}{\xi_2\eta_2}.$ This is why it is so easy not to see the fact that you are really working with two different $\mathbb R^2$ when using the standard basis.

You seem to be confusing the notion of an inner product and a basis.

A (real) inner product on a vector space $V$ is a bilinear function $\langle \cdot | \cdot \rangle\colon V\times V\to {\bf R}$ satisfying the standard axioms of positive definiteness and symmetry. It is this inner product that we use to determine whether a given tuple of vectors is orthogonal or orthonormal, not any basis.

Now, if $V$ is finite dimensional and $(v_1,\ldots, v_n)$ is a basis of $V$, there is a unique inner product such that $(v_1,\ldots, v_n)$ is an orthonormal basis. But the converse is far from truth: given any inner product on a space of dimension at least two, there are infinitely many orthonormal bases.

Now, if you have a space $V$ with a fixed inner product $\langle \cdot | \cdot \rangle_1$, you can take a basis $(v_1,\ldots, v_n)$ of $V$ and find the unique inner product $\langle \cdot | \cdot \rangle_2$ such that $(v_1,\ldots,v_n)$ are orthonormal with respect to it, but unless $(v_1,\ldots,v_n)$ are orthonormal with respect to $\langle \cdot | \cdot \rangle_1$, the two inner products are distinct, so it should come as no surprise that the notions of "orthonormal" differ (precisely because of what I have mentioned in the last paragraph about the inner product being unique given a basis which we want to be orthonormal).

Interestingly, it is actually true that any two inner products have a common orthogonal basis, but nonetheless, the notions of orthogonality do not coincide for distinct inner products, either. In other words, while there are special bases which are orthogonal jointly for both inner products, there are always bases which are orthogonal in one, but not in the other basis (for distinct inner products).