Recently, I got obsessed with working out this story, to the detriment of my mathematical work. Here is a quick crib sheet for relating properties of continued fractions to properties of real quadratic fields. Warning: I haven't checked this against a standard reference, so there may be errors.
Preliminary notation: Continued fractions
It is convenient to convert a continued fraction to a sequence of binary symbols; I've been using red and blue dots, which I'll represent here as $r$ and $b$. For example, $\sqrt{13}=[3,1,1,1,1,6,1,1,1,1,6,\cdots]$; I'll write this as
$$r\ r\ r\ b\ r\ b\ r\ b\ b\ b\ b\ b\ b\ r\ b\ r\ b\ r\ r\ r\ r\ r\ r\ \cdots$$
This makes the continued fraction purely periodic instead of having that peculiar $3$ at the beginning; the period starts again in the middle of that last block of six $r$'s.
Note also that the period of the fraction now appears to be twice what it was; when we get to the old period, we have switched colors. Finally, a sequence which starts with $b$'s should be thought of as a continued fraction for a number in $(0,1)$; the sequence starts with zero copies of $r$.
Preliminary notation: Real quadratic fields
Let $K$ be a quadratic field and let $\Lambda$ be a rank $2$ sublattice of $K$, with $\mathbb{Q} \Lambda =K$. Let $\mathrm{End}(\Lambda)$ be the ring of $\theta \in K$ such that $\theta \Lambda \subseteq \Lambda$. This is an order in $\mathcal{O}_K$.
Let $K$ be a real quadratic field, with fixed embedding $K \to \mathbb{R}$, and write $z \mapsto \overline{z}$ for the Galois symetry of $K$. We'll say that two lattices $\Lambda_1$ and $\Lambda_2$ with CM are "strictly equivalent" if there is an element $\alpha \in K$ such that $\Lambda_1 = \alpha \Lambda_2$ with $\alpha$ and $\overline{\alpha}$ both positive.
Summary of results:
$\bullet$ Take any periodic sequence $a_i$ of $r$'s and $b$'s and turn it into a continued fraction. Let $z$ be the value of that continued fraction. Then $z$ is a quadratic irrational, with $z>0$ and $\overline{z}<0$. We have $a_0=r$ if $z>1$ and $a_0=b$ if $z<1$. Extending the periodicity to negative indices, $a_{-1}=r$ if $\overline{z} < -1$ and $a_{-1} =b$ if $\overline{z} > -1$.
$\bullet$ Switching the colors changes $z$ to $1/z$. Reversing the sequence changes $z$ to $-\overline{z}$.
Strict ideal classes
$\bullet$ Let $K = \mathbb{Q}(z)$ and let $\Lambda = \langle 1, z \rangle$. Shifting the periodic sequence does not change the strict equivalence class of $\Lambda$.
$\bullet$ The above is a bijection between periodic sequences of $r$'s and $b$'s, up to shift, and strict equivalence classes of lattices in real quadratic fields.
$\bullet$ Switching the colors corresponds to multiplying by our lattice by an element of
negative norm. So producing a lattice which is equivalent, but not strictly equivalent.
$\bullet$ Reversing the sequence sends $\Lambda$ to $\overline{\Lambda}$. If $\Lambda$ is a lattice in $K$, and $R = \mathrm{End}(\Lambda)$, then $\Lambda \overline{\Lambda}$ is a strictly principal fractional ideal for $R$. (This is a special property of quadratic fields, which I know of no generalization of in higher degree number fields.) So, with the understanding that we treat a fractional ideal as a fractional ideal for its full endomorphism ring, reversing the sequence sends $\Lambda$ to $\Lambda^{-1}$.
Units
$\bullet$ Let $R$ be the endomorphism ring of $\Lambda$. Let $p/q$ be the convergent obtained by truncating the continued fraction just before the first repetition of the block which contains $a_0$. Let $u=p-qz$. Then $u$ is a unit of $R$ with norm $1$, and is the fundamental generator of the group of such units.
For example, $\langle 1, \sqrt{13} \rangle$ has endomorphism ring $\mathbb{Z}[\sqrt{13}]$. We truncate the above sequence to
$$r\ r\ r\ b\ r\ b\ r\ b\ b\ b\ b\ b\ b\ r\ b\ r\ b$$
or
$$[3,1,1,1,1,6,1,1,1,1] = \frac{649}{180}$$
and $649-180 \sqrt{13}$ is the fundamental positive unit of $\mathbb{Z}[\sqrt{13}]$.
$\bullet$ The color reversal of our sequence is a shift of itself if and only if $R$ has units of norm $-1$. We can recover them by the same recipe, truncating before the color reversed copy of $a_0$.
For example,
$$[3,1,1,1,1] = \frac{18}{5}$$
and $18-5 \sqrt{13}$ is the fundamental unit of norm $-1$ in $\mathbb{Z}[\sqrt{13}]$.
Note, by the way, that we have not yet seen the fundamental unit of $\mathbb{Q}(\sqrt{13})$, which is $(3-\sqrt{13})/2$. That's because $\langle 1, \sqrt{13} \rangle$ doesn't have CM by this unit.
We can obtain the same unit from two quite different looking continued fractions. For example
$$\sqrt{10} = r\ r\ r\ b\ b\ b\ b\ b\ b\ r\ r\ r \cdots \quad \sqrt{10}/2 = r\ b\ r\ b\ b\ r\ b\ r \cdots$$
where I have given a full period for each fraction. Truncating to before the middle block gives
$$r\ r\ r = \frac{3}{1} \quad r\ b\ r\ = \frac{3}{2}.$$
Both of these give the unit $3-\sqrt{10} = 3-2 \frac{\sqrt{10}}{2}$.
Fixing the endomorphism ring; working with triples $(a,b,c)$
$\bullet$ Let $R = \mathbb{Z}[\sqrt{D}]$, for $D>0$ and not square. The continued fractions which give rise to rings containing $R$ correspond to ordered triples $(a,b,c)$ of integers with $D=b^2+ac$ and $a$, $c>0$, by the recipe $(a,b,c) \mapsto (b+\sqrt{D})/a$.
The corresponding ring is exactly $\mathbb{Z}[\sqrt{D}]$ if and only if $(a,2b,c)$ have no common factor.
$\bullet$ Let $R = \mathbb{Z}[(1+\sqrt{D})/2]$ with $D \equiv 1 \mod 4$, positive and not square. The continued fractions which give rise to rings containing $R$ correspond to ordered triples $(a,b,c)$ with $D=b^2+ac$, $a$ and $c>0$, and the additional condition that $b$ is odd and $a$ and $c$ are even.
Example: If we want to get the ring $\mathbb{Z}[(1+\sqrt{13})/2]$, we need to pick $z$ so that $\langle 1, z \rangle$ has CM by this ring. An obvious choice is $z=(1+\sqrt{13})/2$, with $(a,b,c) = (2,1,6)$. The continued fraction is
$$r\ r\ b\ b\ b\ r\ r\ r\ b\ b\ b\ r\ r\ r\ b\ b\ b\ $$
or $[2, 3,3,3,3,\ldots]$ in conventional notation.
$\bullet$ The corresponding ring is exactly $\mathbb{Z}[(1+\sqrt{D})/2]$ if and only if $(a,b,c)$ have no common factor.
$\bullet$ There are only finitely many $(a,b,c)$ for any $R$.
$\bullet$ Adding an $r$ at the beginning of the sequence changes $(a,b,c)$ to $(a,a+b,c-a-2b)$. Adding a $b$ at the beginning changes $(a,b,c)$ to $(a-c-2b,b+c,c)$. Reversing the sequence changes $(a,b,c)$ to $(a,-b,c)$; color switching the sequence sends $(a,b,c)$ to $(c,-b,a)$.
Continued fractions with special symmetry
$\bullet$ A continued fraction is a shift of its color switch if and only if $R$ contains a unit with norm $-1$; we have already described how to find this unit.
$\bullet$ A continued fraction is a shift of its reversal if and only if it is a $2$-torsion class in the strict ideal class group.
Consider continued fractions which equal their reversal, so the periodic sequence starts at the middle of an even block of $r$'s or $b$'s, like the sequence for $\sqrt{13}$ above. These correspond to $z = \sqrt{D}/a$ for some divisor $a$ of $D$.
Consider continued fractions which are off from one by a shift of their reversal, so the periodic sequence starts in the middle of an odd block of $r$'s or $b$'s. These correspond to $z = (b+\sqrt{D})/(2b)$. If $D$ is odd, then we can take $b$ to be any divisor of $D$. If $D$ is $2 \mod 4$, then there are no solutions to $b^2+2bc=D$. If $D$ is $0 \mod 4$, then we can take $b$ of the form $2 b'$, where $b'$ is a divisor of $D/4$.
$\bullet$ Let $(-)$ denote the strict ideal class of principal $\Lambda$ ideals generated by elements of negative norm. A continued fraction is a shift of its color switched reversal if and only if $\Lambda^2 = (-)$ in the strict ideal class group.
A continued fraction actually equals its color switched reversal if and only if $a=c$. In other words, such continued fractions for $R = \mathbb{Z}[\sqrt{D}]$ are in bijection with solutions to $a^2+b^2=D$ with $a$ and $b>0$, and $GCD(a,2b)=1$. Such continued fractions for $R=\mathbb{Z}[(1+\sqrt{D})/2]$ are in bijection with solutions to $a^2+b^2=D$ with $a$, $b>0$, such that $a$ even and $b$ odd.
Example: We have $34=3^2+5^2$. So take $z=(3+\sqrt{34})/5$. Take $\Lambda$ to be the lattice $\langle 1, (3+\sqrt{34})/5 \rangle$, which is strictly equivalent to the ideal $I=\langle 5, 3+\sqrt{34} \rangle$ in $\mathbb{Z}[\sqrt{34}]$. This is a non-principal prime ideal dividing $5$. We have $I^2 = \langle 3+\sqrt{34} \rangle$, which is principal, but not strictly principal.
The corresponding continued fraction is
$$r\ b\ r\ r\ r\ b\ b\ b\ r\ b\ r\ b\ r\ r\ r\ b\ b\ b\ r\ b\ r\ b\ \cdots$$
or $[1,1,3,3,1,1,1,1,3,3,1,1,1,1,\cdots]$ in conventional notation. This sequence is its own color switched reversal, reflecting that $I^2=(-)$. However, it is not a shift of its own reversal, reflecting that $I^2$ is not strictly principal, and it is not a shift of its color switch, reflecting that $\mathbb{Z}[\sqrt{34}]$ does not have a unit of norm $-1$.
Best Answer
See Remark 2.2 here.
If you expand $p/q$ into a continued fraction then the successive convergents, as columns of a $2 \times 2$ matrix, have determinant $\pm 1$. Provided $p/q$ is in reduced form and $q > 0$, the last convergent $p_n/q_n$ in the continued fraction for $p/q$ will have $p_n = p$ and $q_n = q$. Let the second to last convergent be $p_{n-1}/q_{n-1}$. Then $p_{n-1}q_n - q_{n-1}p_n = \pm 1$, and it is easy to pass from such an equation to a $2 \times 2$ integral matrix with determinant $1$ and first column $\binom{p}{q}$.
Let's illustrate these algorithms by starting with the second question on the rational number $p/q = 37/11$, which is in reduced form. What is a matrix in ${\rm SL}_2(\mathbf Z)$ with first column $\binom{37}{11}$? The continued fraction of $37/11$ is $[3,2,1,3]$ and the successive convergents in this continued fraction are $3/1$, $7/2$, $10/3$, and $37/11$. Using the last two convergents, we obtain $\det(\begin{smallmatrix}10&37\\3&11\end{smallmatrix}) = -1$, so $11 \cdot 10 - 37 \cdot 3 = -1$, so $37(3) - 11(10) = 1$. Thus the matrix $A = (\begin{smallmatrix}37&10\\11&3\end{smallmatrix})$ is in ${\rm SL}_2(\mathbf Z)$ with first column $\binom{37}{11}$, so $A(\infty) = 37/11$.
Next, for the matrix $A = (\begin{smallmatrix}37&10\\11&3\end{smallmatrix})$ in ${\rm SL}_2(\mathbf Z)$, how can we write $A$ in terms of $S$ and $T$? For this we will use a continued fraction for the first column ratio $37/11$ using nearest integers from above rather than from below: $37/11 = 4 - 1/(2 - 1/(3 - 1/(2 - 1/2)))$. Using the entries $4, 2, 3, 2$, and $2$, form the matrix product $M = T^4ST^2ST^3ST^2ST^2S$. Its first column will be $\binom{37}{11}$ but its second column might not match that of $A$, so $M^{-1}A$ will be a power of $T$. Indeed, $M = (\begin{smallmatrix}37&-27\\11&-8\end{smallmatrix})$ and $M^{-1}A = (\begin{smallmatrix}1&1\\0&1\end{smallmatrix}) = T$, so $$ A = MT = T^4ST^2ST^3ST^2ST^2ST. $$