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So the answer to your question is: there is no coordinate system in the large with nice properties, unless $M$ is flat.

You see, the question was confused between choosing a gauge and choosing a coordinate system, these are not the same things. If this confusion is straightened out, it gets two different answers: If $M$ is pseudo-Riemannian, yes there exists a choice of gauge which means the metric can be represented by a tensor, not a twisted tensor. But no, there does not exist any prescription for coordinates which have nice properties in the large unless $M$ is flat.

EDIT for the comment by the OP.

The principle of general relativity is that there is no natural way to distinguish between one set of coordinates and another. That is the whole point of GR, its philosophy, if you will. There is no physical criterion to use to say one coordinate system is better than another.

Perhaps you already knew that, so let us consider choices which have no physical motivation or significance but look pretty. E.g., geodetic coordinates. For any $M$ and any given point $p$ you can define local coordinates in a small neighbourhood of $x$ in $M$ which are geodetic in the sense that they nicely describe parallel transport along the coordinate axes. But they have no global significance, they don't do anything for the whole potato, only for the one point $x$, because as soon as you parallel transport something a finite distance away from $x$, what you get depends on the path you took to get there. They have « local » significance, not « global » significance, and the reason there is a difference between local and global is the geometric fact of non-integrability, which is inherent in the curved geometry of $M$. Only if $M$ is flat is the situation « integrable.» In fact, this is the definition of curvature. Curvature is defined as the deviation from integrability of this parallel transport you do in a geodetic coordinate system.

Having now disposed of diff($M$), I briefly state what everyone already knows: for any Riemannian or Lorentzian manifold, there exists a gauge which makes the metric field a tensor, this is explained by Weyl in his book, except he calls it a calibration. So that answers your question for classical GR.