This question is about the definition of order of a section of a bundle at a point, and the related notion of associated divisor.
Let us look at a specific example, the discriminant $\Delta(z)$ on $X(1) = \mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}$. If $q$ is a local parameter at the cusp $i\infty$ then $$ \Delta(q) = q + \sum_{n=2}^\infty a_n q^n,$$ and one would be tempted to deduce
$$\mathrm{ord}_{\,i\infty}(\Delta) = 1.$$
On the other hand one could read the excellent answer of Georges Elencwajg to this question. If $U$ is a small neighborhood around $i\infty$, and we denote by $\mathcal{S}_K$ the bundle of cusp forms of weight $k$ and by $\mathcal{M}_k$ the bundle of modular forms of weight $k$, then $$\mathrm{H}^0(U,\mathcal{S}_k) \simeq \mathrm{H}^0(U,\mathcal{O}_X)\qquad \text{via } f(q) \mapsto f(q)/q,$$ but $$\mathrm{H}^0(U,\mathcal{M}_k) \simeq \mathrm{H}^0(U,\mathcal{O}_X)\qquad \text{via } g(q) \mapsto g(q).$$ Thus I would now say that
$$ \mathrm{ord}_{\,i\infty}(\Delta) = \begin{cases} 0 & \text{ assuming } \Delta \in \mathrm{H}^0(X(1),\mathcal{S}_{12}),\\ 1 & \text{ assuming } \Delta \in \mathrm{H}^0(X(1),\mathcal{M}_{12}),\\ \end{cases} $$
Consequently, I wouldn't have a notion of $\mathrm{div}(\Delta)$ if I don't fix the bundle of whom $\Delta$ is a section.
This looks quite weird to me, and the experts I asked to just told me: "look at the $q$-expansion, since it starts with $q$ it has order $1$".
Summing up:
Is it true that $\mathrm{ord}_{\,i\infty}(\Delta)$ is 0, resp. 1, if I look at it as a section of $\mathcal{S}_{12}$, resp. $\mathcal{M}_{12}$?
Does it imply that I don't have a "universal" notion of divisor of a section of a bundle, as long as I don't specify of which bundle it is a section?
Thank you very much!
As pointed out by David Loeffler in the comments both points $(1)$ and $(2)$ are correct.