My math methods textbook (by Riley, Hobson, and Bence, RHB) gives the explanation attached at the end of this post for how we can move towards simplifying Fourier coefficients in the case of symmetry about the quarter wave position in a periodic function $f(x)$. I don't immediately follow how they arrive at the equation for $b_r$. In particular, if I make the $s=x-L/4$ substitution noted by RHB then I arrive at
$$b_r=\frac{2}{L} \int^{x_0+L}_{x_0}f(x)\sin\bigg(\frac{2\pi rx}{L}\bigg)\mathrm{dx}=\frac{2}{L} \int^{x_0+3L/4}_{x_0-L/4}f(s+L/4)\sin\bigg(\frac{2\pi rs}{L}+\frac{\pi r}{2}\bigg)\mathrm{ds}.$$
It's not clear to me how to then show that this line last equals
$$\frac{2}{L} \int^{x_0+L}_{x_0}f(s)\sin\bigg(\frac{2\pi rs}{L}+\frac{\pi r}{2}\bigg)\mathrm{ds}.$$
$f(s+L/4)$ is periodic, sure, so I can change the integrals bounds to what I need them to be. But I still have $(s+L/4)$ as the argument to $f$ and it seems to me that it would not be permissible to change this to $s$ since the way that interacts with the "modulating phase" of the $\sin$ would change.
As you might have suspected, the explanation in the textbook is confusing. What you have done is correct. $\newcommand{\d}{\,d}$
A minor error
When the textbook describes $f(x)$, "suppose that $f(x)$ has even or odd symmetry about $L/4$, i.e. $f(L/4 − x) =\pm f(x − L/4)$" should have been "... i.e. $f(L/4 - x) =\pm f(L/4+x)$".
A confusing formula
When the following formula is presented in the textbook $$b_r=\frac{2}{L} \int^{x_0+L}_{x_0}f(s)\sin\bigg(\frac{2\pi rs}{L}+\frac{\pi r}{2}\bigg)\d s,\tag{***}\label{***}$$ the $f(\cdot)$ in it is NOT the original function $f(\cdot)$. This is probably the cause of confusion.
For example, let $f(x)=\sin(\frac{2\pi x}L)$. The Fourier coefficient $b_1$ of $f(x)=$ is $1$ (and all other Fourier coefficients of $f(x)$ is $0$). However, had we used the formula $\eqref{***}$ above to compute $b_1$, using $f(s)=f(x)|_{x=s}$, we would have found $b_1=0$.
What the heck is that function $f(s)$ in the formula $\eqref{***}$?
It is the function $g$ defined below. (What happens in the book is an example of using the same letter to denote different functions. If the independent variable is $x$, then $f(x)$ means the given function. If it is $s$, then $f(s)$ means a different function, $g$ below!)
To clear the confusion, let us define function $g:\Bbb R\to \Bbb R$ such that $g(x)=f(x+L/4)$. Thanks to what have been found previously in the textbook, we have the first equality of the following computation.
$$\begin{aligned}b_r &=\frac{2}{L} \int^{x_0+L}_{x_0}f(x)\sin\left(\frac{2\pi rx}L\right)\d x\\ &=\frac{2}{L} \int^{x_0+L}_{x_0}g(x-L/4)\sin\left(\frac{2\pi rx}L\right)\d x\\ &\quad (\text{change of variable, } x=s+L/4)\\ &=\frac{2}{L} \int^{x_0+L-L/4}_{x_0-L/4}g(s)\sin\left(\frac{2\pi r(s+L/4)}L\right)\frac{\d x}{\d s} \d s\\ &\quad (\text{Replace }x_0-L/4\text{ with } x_0\text{ since }L\text{ is a period of the integrand})\\ &=\frac{2}{L}\int^{x_0+L}_{x_0}g(s)\sin\left(\frac{2\pi rs}L+\frac{\pi r}2\right)\d s.\\ \end{aligned}$$
We have obtained again the formula $\eqref{***}$. Note that $f(s)$ in it appears as $g(s)$ here.
The computation above has nothing to do with whether $f(x)$ or $g(x)$ has any symmetry about $L/4$ or not.
Note $g(x)=f(x+L/4)=\pm f(L/4-x)=\pm g(-x)$. That is, if $f(x)$ has even or odd symmetry about $L/4$, then $g(x)$ is an even or odd function respectively.
Now you can read happily the argument latter in section $12.3$ of the textbook, interpreting $f(s)$ as $g(s)$ defined here while reading $f(x)$ as it is.
As a different narrative, we can also say that the formula $\eqref{***}$ is simply an error in the book, where $f(s)$ should be replaced by $f(s+L/4)$. Then we will have to update the references to $f(s)$ latter in the book as well.
Hopefully, I have cleared the situation. Hopefully the next edition of the textbook will rectify the minor error as well as the confusing $f(s)$.