I'm wondering how come between the coordinate system $\xi \eta$, rotated -45° with respect to $x y$ coordinate system, and the $xy$ coordinate system there are the following relationships:
$$x = \frac{\xi + \eta}{\sqrt{2}}$$
$$y = \frac{-\xi + \eta}{\sqrt{2}}$$
Thank you for your time.
I can give you one way to look at at. The coordinate system of the $xy$ axis are determined by the vectors $(1,0)$ and $(0,1)$. What this means is any point $(a,b)$ can be expressed as a linear combination $$\left(\begin{matrix} a \\ b \end{matrix}\right)= a\left(\begin{matrix} 1 \\ 0 \end{matrix}\right) +b\left(\begin{matrix} 0 \\ 1 \end{matrix}\right) = ax+by$$
But what happens to $(1,0)$ and $(0,1)$ after a $-45^o$ rotation? By drawing a picture, using the "special right triangles" where the internal angle is $45^o$, and naming the coordinates after rotation $\xi$ and $\eta$, we see that $$ x = \left(\begin{matrix} 1 \\ 0 \end{matrix}\right) \to \left(\begin{matrix} \frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}} \end{matrix}\right) = \xi $$ and $$ y = \left(\begin{matrix} 0 \\ 1 \end{matrix}\right) \to \left(\begin{matrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{matrix}\right)= \eta $$
We can see the identities in your question applied to $(1,0)$ and $(0,1)$ directly from the 2 equations, since $$ \frac{\xi + \eta}{\sqrt{2}} = \frac{1}{\sqrt{2}} \left[ \left(\begin{matrix} \frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}} \end{matrix}\right) + \left(\begin{matrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{matrix}\right) \right] = \frac{1}{\sqrt{2}} \left(\begin{matrix} \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \end{matrix}\right) = \left(\begin{matrix} \frac{2}{\sqrt{2}\sqrt{2}} \\ 0 \end{matrix}\right) = \left(\begin{matrix} 1 \\ 0 \end{matrix}\right) = x $$ and almost identical calculation for $y$