Time-reversal and delay commutativity and resulting signals
Your counterexample to the book's assertion
is confusing between two different uses for $n$.  There was a question earlier
in which some user (endolith? datageist?) gave an answer containing
a detailed description of what exactly
this confusion is and how to interpret the results correctly. My cursory search
has not found this great answer, and so I will just exhibit
a simple example.

Suppose that 
$x[n] = \begin{cases}1, &amp;n=0,\\2, &amp; n = 1,\\0, &amp;\text{otherwise,}\end{cases}$


Delaying $x[n]$ by one unit gives
$x_d[n] = \begin{cases}1, &amp;n=1,\\2, &amp; n = 2,\\0, &amp;\text{otherwise,}\end{cases}$


and time-reversing this gives
$x_{d,r}[n] = \begin{cases}1, &amp;n=-1,\\2, &amp; n = -2,\\0, &amp;\text{otherwise,}\end{cases}$  which equals $x[-n-1]$ exactly as the book says it does. Don't believe this?
Here is a table of values for the two functions:
$$\begin{array}{|l|c|c|c|c|c|c|c|c|c|}
\hline\\n&amp;\cdots&amp;-3&amp;-2&amp;-1&amp;0&amp;1&amp;2&amp;3&amp;\cdots\\
\hline\\
x[n]&amp;0&amp;0&amp;0&amp;0&amp;1&amp;2&amp;0&amp;0&amp;0\\
\hline\\x_{d,r}[n]&amp;0&amp;0&amp;2&amp;1&amp;0&amp;0&amp;0&amp;0&amp;0\\
\hline\\
-n-1&amp;\cdots&amp;2&amp;1&amp;0&amp;-1&amp;-2&amp;-3&amp;-4&amp;\cdots\\
\hline\\
x[-n-1]&amp;\cdots&amp;0&amp;2&amp;1&amp;0&amp;0&amp;0&amp;0\\
\hline\\
\end{array}$$

OK, thus far?  On  the other hand,


Time-reversing $x[n]$ gives
$x_r[n] = \begin{cases}1, &amp;n=0,\\2, &amp; n = -1,\\0, &amp;\text{otherwise,}\end{cases}$
and delaying this by one time unit gives
$x_{r,d}[n] = \begin{cases}2, &amp; n = 0,\\1, &amp;n=1,\\0, &amp;\text{otherwise,}\end{cases}$
which equals $x[-n+1]$ exactly as the book says it does. Don't believe this either? Here
is the table for it. 
$$\begin{array}{|l|c|c|c|c|c|c|c|c|c|}
\hline\\n&amp;\cdots&amp;-3&amp;-2&amp;-1&amp;0&amp;1&amp;2&amp;3&amp;\cdots\\
\hline\\
x[n]&amp;0&amp;0&amp;0&amp;0&amp;1&amp;2&amp;0&amp;0&amp;0\\
\hline\\x_{r,d}[n]&amp;0&amp;0&amp;0&amp;0&amp;2&amp;1&amp;0&amp;0&amp;0\\
\hline\\
-n+1&amp;\cdots&amp;4&amp;3&amp;2&amp;1&amp;0&amp;1&amp;2&amp;\cdots\\
\hline\\
x[-n+1]&amp;0&amp;0&amp;0&amp;0&amp;2&amp;1&amp;0&amp;0&amp;0\\
\hline\\
\end{array}$$






So how can we establish these results without setting up tables?

Well, $x_d$ is a sequence whose $n$-th term $x_d[n]$ equals $x[n-n_0]$
for all choices of $n$; in other words, take whatever is inside the square brackets
after $x_d$, subtract $n_0$ from it, and stick it in as the argument/index
for $x$. Now, for
any sequence $y$, its time-reversal $y_r$ is a sequence for which
$y_r[k] = y[-k]$ for all choices of $k$ (negate the argument of $y_r$ and
stick it in as the argument for $y$. Thus, the time-reversal of
$x_d$ is given by

$$x_{d,r}[k] = (x_d)_r[k] = x_d[-k] = x[(-k)-n_0] = x[-k-n_0]$$
as the book says, and not $x[-k+n_0]$ as the OP claims it should be.

Similarly, $x_r[n] = x[-n]$ and so
$$x_{r,d}[k] = (x_r)_d[k] = x_r[k-n_0] = x[-k+n_0]$$
as the book says it should be, and not $x[-k-n_0]$ as the OP
claims it should be.