This is all about how to make consistent choices of square roots. If you want to attach a meaning to half-integer weight modular forms of level $\Gamma$, then you need to pick, in some consistent fashion, a square root of $c\tau + d$, for every $\tau \in \mathcal{H}$ and every $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ in your group $\Gamma$. If $\Gamma \subseteq \Gamma_0(4)$, then the transformation law for the Jacobi theta function shows that there's a consistent set of choices, and that leads to the "usual" definitions.
So the question is: if $\Gamma$ isn't contained in $\Gamma_0(4)$, how do we choose the square roots? There's a gadget called the metaplectic group $M$ which is cooked up specifically to understand what this means: an element of $M$ consists of a pair $(\gamma, j_\gamma)$, where $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is an element of of $SL_2(\mathbb{R})$, and $j_\gamma$ is a holomorphic function $\mathcal{H} \to \mathbb{C}$ such that $j_\gamma(\tau)^2 = c\tau + d$. The group law in $M$ is a bit messy to write down, although it's very natural; it's rigged so that
- Forgetting $j_\gamma$ gives a group homomorphism $M \to SL_2(\mathbb{R})$ with kernel $\pm 1$.
- For any $k \in \tfrac{1}{2}\mathbb{Z}$, there's a "weight $k$" right action of $M$ on holomorphic functions $\mathcal{H} \to \mathbb{C}$, given by
$$\left[f \mid (\gamma, j_\gamma)\right](\tau) = j_\gamma(\tau)^{-2k} f(\gamma \cdot \tau).$$
If $k \in \mathbb{Z}$ then this action factors through the map $M \to \operatorname{SL}_2(\mathbb{R})$ and it's just the usual weight $k$ action; if $k$ is a half-integer, then it really matters which $j_\gamma$ you pick.
So it's obvious how to define half-integer weight modular forms for every discrete subgroup $G$ of $M$.
(If you care about Lie groups, you might be interested to know that $\operatorname{SL}_2(\mathbb{R})$ is homotopy equivalent to the circle group $SO_2(\mathbb{R})$ so its fundamental group is $\mathbb{Z}$. Thus it has a unique degree $n$ covering for every $n \ge 1$. It turns out that these coverings are also Lie groups, and $M$ is the $n=2$ case of this construction. This $M$ is an example of a real Lie group having no finite-dimensional faithful representation.)
Now here's the point: the exact sequence
$$ 0 \to \{ \pm 1\} \to M \to SL_2(\mathbb{R}) \to 0$$
splits over $\Gamma_0(4)$, i.e. there's a homomorphism $\Gamma_0(4) \to M$ whose composite with the projection $M \to SL_2(\mathbb{R})$ is the identity map of $\Gamma_0(4)$. (This homomorphism is exactly the choice of square root in the transformation law for $\theta$.) Hence any subgroup of $\Gamma_0(4)$ can be regarded as a subgroup of $M$ in a natural way. However, if you have some more general subgroup of $SL_2(\mathbb{Z})$, you have to choose how you want to lift it to $M$, and there might not even exist such a lifting (if I remember correctly there is an obstruction coming from some class in an $H^2$).
In some sense the "right" notion of level for a half-integer weight form is a subgroup of $M$, not of $\operatorname{SL}_2(\mathbb{R})$; the existence of the splitting over $\Gamma_0(4)$ is somehow an "accident", and trying to make the definitions work for more general subgroups of $\operatorname{SL}_2(\mathbb{Z})$ is a morally unsound question.
