Relationship between $\star$-products in phase-space QM and NC geometry

What exactly is the relationship between $$\star$$-products in phase-space quantum mechanics, i.e.

$$(f \star g) (x,p) = f(x,p) e^{\frac{i \hbar}{2} ( \overleftarrow{\partial_x} \cdot \overrightarrow{\partial_p} - \overleftarrow{\partial_p} \cdot \overrightarrow{\partial_x})} g(x,p) ~,$$ and $$\star$$-products in $$2n$$-dimensional non-commutative spaces, i.e.,

$$(f \star g) (X) = \left . e^{\frac{i\hbar}{2} \theta^{\mu\nu} \frac{\partial}{\partial X^\mu}\frac{\partial}{\partial Y^\nu}} f(X) g(Y)\right \vert_{Y \to X} ~~,$$ where $$\theta^{\mu\nu}$$ is a constant anti-symmetric tensor?

To be a bit more specific, is there a way to reinterpret/reformulate the phase-space non-commutativity as spatial non-commutative geometry? Or something of that sort.

If the phase space is $$\mathbb{R}^{2r+k}$$, and $$\theta^{\mu\nu}$$ has rank $$2r$$ then there exists a bijective linear coordinate transformation that brings the anti-symmetric real matrix $$\theta^{\mu\nu}$$ on standard form $$\theta^{\prime \mu\nu}~=~\begin{pmatrix} \mathbb{0}_{r\times r} & \mathbb{1}_{r\times r}& \cr -\mathbb{1}_{r\times r} & \mathbb{0}_{r\times r}& \cr && \mathbb{0}_{k\times k} \end{pmatrix}.$$ If furthermore $$\theta^{\mu\nu}$$ is non-degenerated (that is: $$k=0$$) then its $$\star$$-product is equivalent to the standard Groenewold-Moyal $$\star$$-product in Darboux/canonical coordinates.
Finally, we should stress that there are important conceptional issues to be sort out if time $$x^0$$ is non-commutative. In contrast, this does not happen in QM, where time is a parameter (as opposed to an operator).
The classic Groenewold star product $$(f \star g) (x,p) = f(x,p)~ e^{\frac{i \hbar}{2} ( \overleftarrow{\partial_x} \cdot \overrightarrow{\partial_p} - \overleftarrow{\partial_p} \cdot \overrightarrow{\partial_x})} g(x,p)=\left . e^{\frac{i \hbar}{2} ( \partial_x \cdot \partial_{p'} - \partial_p \cdot \partial_{x'})} f(x,p)g(x',p') \right \vert_{x'=x,~ p'=p}$$ is routinely recast as $$(f \star g) (X) =\left . e^{\frac{i\hbar}{2} \epsilon^{\mu\nu} \frac{\partial}{\partial X^\mu} \frac{\partial} {\partial Y^\nu}} f(X) g(Y) \right \vert_{Y \to X}\\ X^\mu\equiv (x,p) ~.$$
It should be apparent how to generalize this to multidimensional phase spaces, with r space coordinates, wrinkle their geometry, and append degenerate dimensions, as the proper answer details. You may be suggesting this with the dots dotting r x components to r p components, as with the symbol used, in which the expression is already generalized to higher r. That is, you may give $$X^\mu$$ another index, j, ranging from 1 to r, in a direct product tensor space, so $$X^\mu \to X^\mu_j$$, in which case the exponent becomes $$\frac{i\hbar}{2} \epsilon^{\mu\nu} \frac{\partial}{\partial X^\mu_j } \frac{\partial} {\partial Y^\nu_j}$$; and then, finally, omit the j indices, understanding a dot product in their space. In your conventions, if you really mean the dot in your starting expression, just write $$\frac{i\hbar}{2} \epsilon^{\mu\nu} \frac{\partial}{\partial X^\mu }\cdot \frac{\partial} {\partial Y^\nu}$$. The fine real answer I am deferring to blends the j indices into an extension of $$\mu,\nu,...$$ to those of a 2r-dimensional symplectic vector.