Let $\Gamma_1(4)$ be the principal subgroup $\bigg\{\gamma\in\text{SL}_2(\mathbb{Z})\colon\gamma\equiv\begin{pmatrix}1&b\\0&1\end{pmatrix}\mod(4)\bigg\}$.
I am trying to show that the cusps of $\Gamma_1(4)$ are the orbits of $0,\frac 12,\infty$.
I have almost completed this, but I want to show that all $\frac{r}{s}$ with $\gcd(r,s)=1$ and $s\equiv 2\mod 4$ are all in the orbit $\Gamma_{1/2}:=\Gamma_1(4)\cdot\frac 12$. I wanted to do this as follows. Since $\begin{pmatrix}1&1\\ 0&1\end{pmatrix}\in\Gamma_1(4)$, there exists $k\in\mathbb{Z}$ such that $r+2kt\in\{-t,t\}$, so $\begin{pmatrix}1&1\\ 0&1\end{pmatrix}^k\dfrac {r}{2t}=\dfrac{r+2kt}{2t}=\pm\dfrac{1}{2}$.
Is this approach correct?
It feels incorrect since I only use one of the two generators of $\Gamma_1(4)$.
The approach in the question is not correct. Let $\frac{r}{2t}$ be given, with $\gcd(r,2t)=1$ and $t$ odd. Then there exists a $k\in\mathbb{Z}$ such that $-t<r+2kt\leq t$, which is different from $r+2kt\in\{-t,t\}$. Note that $r+2kt$ is odd.
So indeed, it is necessary to use another generator of $\Gamma_1(4)$, for example use the matrix $\begin{pmatrix}1&0\\4l&1\end{pmatrix}$, where there exists a $l\in\mathbb{Z}$ such that $-2r<4lr+2t\leq 2r$. Note that $4lr+2t\equiv 2\mod 4$.
These two steps give an algorithm: if $\frac{r}{2t}\leq-1/2$ or $\frac{r}{2t}>1/2$, use the first step; if $-1/2<\frac{r}{2t}<1/2$, use the second step.
This algorithm terminates in a finite number of steps, because in each step, the absolute value of either the numerator or denominator is decreased (except for $\frac{r}{2t}=\frac{1}{2}$, but this is the only exception).