Why does the star product satisfy the "Bopp Shift relations": $f(x,p)star g(x,p)=f(x+frac{i}{2}partial_p,p-frac{i}{2}partial_x) g(x,p)$?

In (Curtright, Fairlie, Zachos 2014), the authors mention (Eq. (14) in this online version) the following relation, known as "Bopp shifts":
$$f(x,p)star g(x,p)=fleft(x+frac{i}{2}partial_p,p-frac{i}{2}partial_xright) g(x,p),tag1$$
where the $star$-product is defined as
$$starequivexpleft[ frac{i}{2}(partial_x^Lpartial_p^R – partial_p^L partial_x^R)right],tag2$$
and I’m denoting with $partial_i^L,partial_i^R$ the partial derivative $partial_i$ applied to the left or right, respectively.

I’m trying to get a better understanding of where this comes from. As far as I understand, $f$ and $g$ are regular functions here (usually Wigner functions I suppose), so $fstar g$ should produce another "regular" function.
If this is so, what do the derivatives in the argument of $f$ mean exactly?
If I were to simply apply (2) to $fstar g$, I would naively get the following expression:
$$fstar g =
sum_{s=0}^infty frac{(i/2)^s}{s!} sum_{k=0}^s (-1)^k
(partial_x^{s-k}partial_p^k f) (partial_x^k partial_p^{s-k}g). tag3
$$
How is this compatible with (1)?

In fairness, if I were to very handwavily apply (2) to (1) without being too careful, I could think of $partial_p^R$ and $partial_x^R$ in the exponential as $c$ numbers, and the $star$ operator as only acting on $f$, so that $frac{i}{2}partial_x^Lpartial_p^R$ would be the operator enacting the translation $xmapsto x+frac{i}{2}partial_p$, and similarly for the other term in the exponential. At a purely formal level this seems to make sense, but more concretely I’m not sure what the expression (1) is even supposed to represent, and how it is consistent with (3).