Note: In the course of writing up this question, I realized the answer. In particular, what the author was saying went over my head because the notation was unusual to me. In the course of changing the notation to make it readable to anyone who might answer the question, I understood finally what the author was saying. In order to make this not a complete waste of time, I am typing up my realization as a community wiki answer so others can avoiding wasting their time the same way.
Basic Idea: This is sort of a "extreme value version of inverse transform sampling". In particular, for any continuous distribution function $F$, the random variable $\Xi$ has the exact same distribution.
Details: In particular, the distribution of $\Xi$ can be seen to be related to the distribution of the maximum of $n$ i.i.d. uniform random variables. More importantly the distribution of $\Xi$ has a simple closed-form density function and therefore moments and other quantities of $\Xi$ are simple to calculate. Thus writing $\max_{1 \le i \le n}X_i$ in terms of $\Xi$ makes at least the moments of $\max_{1 \le i \le n}X_i$ possibly easier to calculate, and other quantities as well in some cases.
Regarding the $\operatorname{Unif}(a,b)$ example I mentioned earlier, the author shows that
$$ \max_{1 \le i \le n }X_i = b - \frac{b-a}{n} \Xi \,. $$
The claim I didn't understand was why this implied that:
$$ \mathbb{E} \left[ \max_{1 \le i \le n } X_i \right] = b - \frac{n}{n+1}(b-a)\,, $$
i.e. why one has that
$$\mathbb{E} [ \Xi] = \frac{n^2}{n+1} \,. $$
However, it is easy to see with basic calculus that:
$$\mathbb{E} [ \Xi] = \frac{1}{n^{n-1}} \int_0^n (\xi)(\xi^{n-1}) d \xi = \frac{n^2}{n+1} \,. $$
It is obvious to me now as written, but the notation used in the book was originally unclear to me.