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The subtraction is well defined, at least, because the $g$'s form a group under addition (they're not elements of the gauge group, they're metrics). It seems like this definition of the FD determinant differs from the convention of others; I've only seen it where the argument of the FD determinant is subjected to the gauge transformation in the integral, i.e. $\Delta[g]^{-1} = \int \mathcal D \xi \delta(g^\xi - g_0)$
This whole example is more of a schematic to figure out the general idea of what goes on when you accelerate a charge. If you actually do the calculation (see these formulas) you see that field at some reference point in fact depends on a lot of "arbitrary" points, namely all of the points in its light cone.
Put dollar signs around your formula to make it render inline (like $x = 2$) or two dollar signs on each side to put the formula on its own line. Also if you do like the answer don't forgot to mark it as accepted!
