Well you seem to have worked it out but I wrote most of this before your comment happened: I claim there isn't any material difference between your "Minkowski space Fourier transform" and the usual Fourier transform on ${\mathbb R}^n$: in fact write $$ \hat f(\xi)\equiv \int e^{i\eta(x,\xi)} f(x) dx $$ for *any* symmetric non-degenerate bilinear form $\eta$. Then there exists another such form $\eta^{-1}$ so $\eta(x,\eta^{-1}\zeta)= \langle x,\zeta\rangle$ where $\zeta \in ({\mathbb R}^n)^\star$. Clearly $$ \hat f(\eta^{-1} \zeta)=({\mathcal F}f)(\zeta)\,,$$ where $\mathcal F$ is the usual --- ``Euclidean'' --- Fourier transform.

In physics texts this $\zeta$ variable is the down-index momentum ($k_\mu$ in e.g. Peskin and Schroeder) while $\xi$ is the up-index momentum ($k^\mu$ in e.g. Peskin and Schroeder). Derivatives play perfectly nice with up/down index notation, which allows one to be blase about whether the Fourier transform involves the up- or down-index $k$.

Mathematically speaking you're taking the Pontryagin dual of Minkowski space seen as a group of translations, which is exactly the same as that of the corresponding Euclidean space. More abstractly the conserved charge in the sense of Noether's theorem corresponding to a translation is the down-index momentum rather than the up-index one.