I have been trying to do this by finding the $\Gamma_0(4)$ orbit of $\infty$, then finding an element not in here (namely $0$) and considering the orbit of this. But I feel like there is a slight error in my computation as I suspect there are in fact 3 non-equivalent cusps.
Here is what I have done thus far:
$\Gamma_0(4)\cdot\infty = \{\left( \begin{array}{cc} a & b \\ 4c & d \end{array} \right)\infty : ad - 4bc = 1\} = \{\frac{a}{4c} : gcd(a,4c) = 1\} = \{\frac{p}{q} : gcd(p, q) = 1, 4|q\}$
and then similarly for $0$. This method would give just the two inequivalent cusps which I am pretty sure is wrong. I feel like the error is probably in my final equality but I cannot see why this would be incorrect.
I guess you know how to compute the number of inequivalent cusps of $\Gamma_{0}(N)$: $\sum_{d\mid N}\phi(\gcd(d,N/d))$. So the number of cusps of $\Gamma_{0}(4)$ has three inequivalent cusps. We know that the cusps $\infty$ and 0 are inequivalent. I claim that $1/2$ is not equivalent to neither $\infty$ nor $0$.
If $1/2$ were equivalent to $0$, then there would exist $j\in\mathbb{Z}$ such that $2j+1\equiv 0\pmod{4}$ (please see p. 99 of 'A first course in modular forms' written by Diamond and Shurman), which is absurd. We conclude that $1/2$ is not equivalent to $0$. Similarly, $1/2$ is not equivalent to $\infty$.