Do derivatives of operators act on the operator itself or are they "added to the tail" of operators?

If we leave out various subtleties related to operators, the core of OP's question (v4) seems to boil down to the following.

What is meant by $$\tag{0}\frac{d}{dx}f(x)?$$ Do we mean the derivative $$\tag{1} f^{\prime}(x),$$ or do we mean the first-order differential operator that can be re-written in normal-ordered$^1$ form as $$\tag{2} f^{\prime}(x)+f(x)\frac{d}{dx}?$$

The answer is: It depends on context. Different authors mean different things. One would have to trace carefully the author's definitions to know for sure. However, if it is written as $\frac{df(x)}{dx}$ instead, it always means $f^{\prime}(x)$, or equivalently, $[\frac{d}{dx},f(x)]$.

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$^1$ A differential operator is by definition normal-ordered, if all derivatives in each term are ordered to the right.