At the start we have (for $m=0$ and $d=1$) :
$$\sqrt{S}=\frac{\sqrt{S}+m}d=a+\frac{\sqrt{S}+m-da}d$$
(with $a,\ m$ and $d$ are integers)
Suppose that $\ d$ divides $(S-m^2)\ $ (this is true for $d=1$ of course).
The fractional part $\displaystyle \frac{\sqrt{S}+m-da}d$ becomes :
$$\frac{\sqrt{S}-da+m}d=\frac{S-(da-m)^2}{d\bigl(\sqrt{S}+da-m\bigr)}$$
The numerator $\ S-(da-m)^2=(S-m^2)+da(2m-da)$ will be divisible by $d$ (from our hypothesis).
If we note $\ m':=da-m\ $ then the numerator divided by $d$ becomes $\ d':=\dfrac{S-(da-m)^2}d=\dfrac{S-m'^2}d$
and the next term to examine will be :
$$\frac{\sqrt{S}+da-m}{\frac{S-(da-m)^2}d}=\frac{\sqrt{S}+m'}{d'}$$
But the conditions are the same as at the start :
$\ d'$ divides $(S-m'^2)\ $ (the fraction is the previous $d$ !) and we may continue our rewriting :
$$\frac{\sqrt{S}+m'}{d'}=a'+\frac{\sqrt{S}+m'-d'a'}{d'}$$
This recurrence shows that these conditions will hold at each iteration.
(To be complete let's add that at each step $\ a:=\left\lfloor\dfrac{\sqrt{S}+m}d\right\rfloor$)
Bounding the error.
The error between a continued fraction $[a_0;a_1,a_2,\ldots]$ and its truncation to the rational number $[a_0;a_1,a_2,\ldots,a_n]$ is given by
$$
|[a_0;a_1,a_2,a_3,\ldots] - [a_0;a_1,a_2,\ldots,a_n]|=\left|\left(a_0+\frac{1}{[a_1;a_2,a_3,\ldots]}\right) - \left(a_0 + \frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)\right|=\left|\frac{[a_1;a_2,a_3,\ldots,a_n]-[a_1;a_2,a_3,\ldots]}{[a_1;a_2,a_3,\ldots]\cdot[a_1;a_2,a_3,\ldots,a_n]}\right| \le \frac{\left|[a_1;a_2,a_3,\ldots,a_n]-[a_1;a_2,a_3,\ldots]\right|}{a_1^2},
$$
terminating with $\left|[a_0;a_1,a_2,a_3,\ldots] - [a_0;]\right|\le 1/a_1$; by iterating this recursive bound we conclude that
$$
\left|[a_0;a_1,a_2,a_3,\ldots] - [a_0;a_1,a_2,\ldots,a_n]\right| \le \frac{1}{a_1^2 a_2^2 \cdots a_n^2}\cdot \frac{1}{a_{n+1}}.
$$
Let $D([a_0;a_1,a_2,\ldots,a_n])$ be the denominator of the truncation $[a_0;a_1,a_2,\ldots,a_n]$ (in lowest terms). Then we have a Liouville number if for any $\mu > 0$, the inequality
$$
a_{n+1} \ge \frac{D([a_0;a_1,a_2,\ldots,a_n])^\mu}{a_1^2 a_2^2 \cdots a_n^2}
$$
holds for some $n$. To give a more explicit expression, we need to bound the growth of $D$.
Bounding the denominator.
Let $D(x)$ and $N(x)$ denote the denominator and numerator of a rational number $x$ in lowest terms. Then
$$
D([a_0;a_1,a_2,\ldots, a_n])=D\left(a_0+\frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)\\ =D\left(\frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)=N([a_1;a_2,a_3,\ldots,a_n]),
$$
and
$$
N([a_0;a_1,a_2,\ldots, a_n])=N\left(a_0+\frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)\\ =N\left(a_0+\frac{D([a_1;a_2,a_3,\ldots,a_n])}{N([a_1;a_2,a_3,\ldots,a_n])}\right) = a_0 N([a_1;a_2,a_3,\ldots,a_n]) + D([a_1;a_2,a_3,\ldots,a_n]) \\ = a_0 D([a_0;a_1,a_2,\ldots, a_n]) + D([a_1;a_2,a_3,\ldots,a_n]).
$$
So
$$
D([a_0;a_1,a_2,\ldots,a_n]) = a_1 D([a_1;a_2,a_3,\ldots,a_n]) + D([a_2;a_3,a_4,\ldots,a_n]),
$$
and the recursion terminates with $D([a_0;])=1$ and $D([a_0;a_1])=D(a_0+1/a_1)=a_1$. Since we have $D([a_0;a_1,a_2,\ldots,a_n]) \ge a_1 D([a_1;a_2,a_3,\ldots,a_n])$, we can say that $D([a_1;a_2,a_3\ldots,a_n]) \le \frac{1}{a_1}D([a_0;a_1,a_2,\ldots,a_n])$, and so
$$
D([a_0;a_1,a_2,\ldots,a_n]) \le \left(a_1 +\frac{1}{a_2}\right) D([a_1;a_2,a_3,\ldots,a_n]) \le (a_1 + 1)D([a_1;a_2,a_3,\ldots,a_n]).
$$
An explicit bound on the size of the denominator is therefore
$$
D([a_0;a_1,a_2,\ldots,a_n]) \le (a_1+1)(a_2+1)\cdots(a_n+1).
$$
Conclusion.
We conclude the following theorem:
The continued fraction $[a_0;a_1,a_2,\ldots]$ is a Liouville number if, for any $\mu > 0$, there is some index $n$ such that $$a_{n+1} \ge \prod_{i=1}^{n}\frac{(a_i + 1)^{\mu}}{a_i^2}.$$
Best Answer
I'm going ahead and summing up what was said in the comments.
To your 1. point, we all agree that you are completely correct that $[a ; a, a, a, \ldots]$ is a root of the $x^{2}-a x-1$ polynomial, and not $x^{2}-a x+1$. So indeed, there must be a typo in the source document.
About your 2. point, and assuming $a>1$, it is easy to verify that if some $\alpha$ verifies $$ \require{cancel} \alpha-1=a-2+\cfrac{1}{1+\cfrac{1}{\alpha-1}} \tag{1} $$ then it follows that $$ \begin{aligned} \alpha-1 &= a-2+\frac{\alpha-1}{\alpha} \\ \alpha^2-\cancel{\alpha} &= a\alpha-\cancel{2\alpha}+\cancel{\alpha}-1 \\ \alpha^2 - a\alpha + 1 &= 0 \end{aligned} $$ $\alpha$ is a root of $x^{2}-a x+1$. Consequently it can be written in continued fraction form as $\alpha = \left[a-1;\overline{1,a-2}\right]$. Also, mon cher ami François Viète assures me that the other root of the polynomial verifies $\alpha\beta=1$, and is therefore $\beta = \left[0; a-1,\overline{1,a-2}\right]$.
As for the $[a ; -a, a, -a, \ldots]$ continued fraction suggested as a possible root of $x^{2}-a x+1$, I can't be sure, but if it is one, then it must be equivalent to either $\alpha$ or $\beta$.
In answer to OP's query about how I arrived at $(1)$: it was essentially a form of reverse-engineering. As I mentioned in the comments, I had found a pattern while playing with such polynomials on WolframAlpha, suggesting the continued fraction form $\left[a-1;\overline{1,a-2}\right]$, so I tried to build the expression from there. Notice that you can add $-1$ on both sides of $\alpha = \left[a-1;\overline{1,a-2}\right]$ and obtain $\alpha-1 = \left[a-2;\overline{1,a-2}\right]$, which allows to write the recursive fraction $(1)$. I verified algebraically that the resulting $\alpha$ was indeed a root of the considered polynomial, and voila.