(Sorry about the generation error!)
To attempt to answer the actual question, I think that it’s a largely two-fold consequence of the historical inertia mentioned in the comments by @coffeemath. On the one hand, it’s simple classroom inertia: it’s ‘always’ been done this way, so we do it this way. On the other hand it’s the practical consideration that since the radical notation does survive in real-world use, students need to learn how to deal with it. None of this, however, justifies the requirement that students deal with it directly, rather than by translating it into a less cumbersome, more easily manipulated notation. Indeed, in my view this is a good occasion to make the point that well-chosen notation makes our mathematical lives easier.
You have put your finger precisely on the statement that is incorrect.
There are two competing conventions with regard to rational exponents.
The first convention is to define the symbol $a^x$ for $a > 0$ only. The symbol $\sqrt[n]{a}$ is defined for negative values of $a$ so long as $n$ is odd, but according to this convention, one wouldn't write $a^{1/n}$, for instance.
In defining $a^{p/q}$ to be $(\sqrt[q]{a})^p$, the author you quoted chose the fraction $p/q$ to be in lowest form so that the definition would be unambiguous. For example, $a^{10/15}$ is defined to be $(\sqrt[3]{a})^2$. However, it is preferable to define $a^{p/q}$ to be $(\sqrt[q]{a})^p$ in all cases and to prove that this definition is independent of the particular representation chosen for $p/q$; this is what more rigorous books tend to do. That is, you prove that if $p/q = r/s$, then $(\sqrt[q]{a})^p = (\sqrt[s]{a})^r$. There is no mention of lowest form.
The competing convention is to also allow $a^x$ to be defined for all $a \ne 0$ and all rational numbers $x = p/q$ that have at least one representation with an odd denominator. You then prove that $(\sqrt[q]{a})^p$ is independent of the particular representation $p/q$ chosen, so long as the denominator is odd. Thus you can write $a^{3/5} = (\sqrt[5]{a})^3 = (\sqrt[15]{a})^{9} = a^{9/15}$. All of that is fine. However, you cannot write $a^{6/10} = (\sqrt[10]{a})^6$, or even $a^{6/10} = \sqrt[10]{a^6}$. The number $a^{6/10}$ is well-defined, but to write down its definition, you must first select a fraction equivalent to $6/10$ that has an odd denominator, which could be $3/5$ or $9/15$ or something else. For $a^{1/2}$, this can't be done at all, so $a^{1/2}$ is undefined for $a < 0$.
The rules for exponents break down if you start allowing $a < 0$ and exponents that can't be written with an odd denominator. For example, the rule $a^{xy} = (a^x)^y$ is valid, but only so long as $x$ and $y$ are both rational numbers that can be written with an odd denominator. This is not the case if you write $a^1 = (a^2)^{1/2}$, despite the fact that both sides of the equation are defined since $a^2 > 0$.
Edit Reading the paper by Tirosh and Even, I was surprised to learn this matter has drawn serious attention from math educators.
A long time ago, I assumed that, apart from complex extensions, $a^x$ for non-integer $x$ should be defined only for $a > 0$. I reasoned that it made no sense to have a function $(-2)^x$ defined only for rational numbers $x$ with odd denominator. I objected strenuously to notations like $(-8)^{1/3}$.
But that was before I taught a calculus class, which is when I realized why some textbook authors are so happy to define $a^x$ for $a < 0$, following the second convention. The reason is that the formula $\frac{d}{dx}(x^r) = rx^{r-1}$ is perfectly valid for $x < 0$ and $r$ with odd denominator.
Best Answer
Let $x, y$ be not negative, $a=1$ and $b$ odd. $$y=x^{\frac{1}{b}} \overset{def.}\Leftrightarrow y^b=x \Leftrightarrow y= \sqrt[b]{x}.$$
Now let $x, y$ not negative, $a$ be arbitrary and $b$ odd. $$y=x^{\frac{a}{b}} \overset{def.}\Leftrightarrow y^b = x^a \Leftrightarrow \sqrt[b]{y^b} = \sqrt[b]{x^a} \Leftrightarrow y = \sqrt[b]{x^a} $$
Since $(\sqrt[b]{x^a})^b = x^a$ and by your proof for integer exponents $((\sqrt[b]{x})^a)^b = (\sqrt[b]{x})^{ab} = x^a$, we have $(\sqrt[b]{x^a})=(\sqrt[b]{x})^a$.
Now let $x, y$ be not negative and $b, d$ odd. $$x^{\frac{a}{b}}x^{\frac{c}{d}}=x^{\frac{ad}{bd}}x^{\frac{bc}{bd}} = (\sqrt[bd]{x})^{ad} (\sqrt[bd]{x})^{bc} \overset{\text{form the integer result}}=(\sqrt[bd]{x})^{ad+bc}= x^{\frac{ad+bc}{bd}}.$$ For even denominators you basically do the same, but pay attention because even roots are not unique ($x^2 = 2 \Rightarrow x= \sqrt{2}$ or $- \sqrt{2}$).
Also note that you have to place proper restrictions on the domain of $x$ and $y$, as for example you might not want to have $\sqrt{-1}$.