Let's say I am trying to find the commutator of operators $\mathbf{A}$ and $\mathbf{B}$, and I get $$[\mathbf{A},\mathbf{B}]=\nabla^2 f(x,y,z).\tag{0}$$
There seems to be some ambiguity here.
In operator notation, which usually does not include the test function $g$, what is the notational standard used to distinguish
$$[\mathbf{A},\mathbf{B}] g = \nabla^2 (f(x,y,z)g)\tag{1}$$
and
$$[\mathbf{A},\mathbf{B}] g = (\nabla^2 f(x,y,z))g\tag{2}$$
where $g$ depends on $x$, $y$, and $z$?
If operators $A$ and $B$ have commutation relation $$[A,B] = g \mathbb I $$ for some function $g:\mathbb R^3 \rightarrow \mathbb C$, then that means $[A,B]\psi = g\psi$, which we might also write $$\bigg([A,B]\psi\bigg)(x,y,z) = g(x,y,z) \psi(x,y,z)$$ My interpretation of OP's expression $(0)$ is precisely this, with $g(x,y,z)\equiv \nabla^2 f(x,y,z)$. However, as OP points out, there is some ambiguity here, so some subsequent clarification on the part of the writer would be warranted. A less ambiguous notation for OP's expressions $(1)$ and $(2)$ might be
$$[A,B] = (\nabla^2 f)\mathbb I + 2(\nabla f) \cdot \nabla + f \nabla^2 \tag{1'}$$ $$[A,B] = (\nabla^2 f) \mathbb I \tag{2'}$$
As is often the case, you need to find the proper balance between notational brevity and clarity which suits your audience and style.
If the former solution was what you intended to mean, then you wouldn't write $d/dx f(x)$ -- to my eyes, I would not think from this expression that the derivative is supposed to act on the test function. Instead, you would keep the product rule in mind and write $$ \frac{df}{dx} + f(x) \frac{d}{dx} $$
