I come from the world of precise satellite orbit determination, so my references also come from there. I realize this means most of the equations in my references relate to orbits and not general measurements, however, I do think you'll find them interesting.
I think a good reference is Fundamentals of Astrodynamics and Applications, by D.A. Vallado. The relevant section is Section 3.7 and onward.
Another good reference is Satellite Orbits, by Montebruck and Gill. Relevant sections are in Chapter 3 (could not find a link).
To summarize: quasi-inertial reference frames are frames such that Newton's laws and special relativity can be applied without any significant impact on accuracy. Usually, for applications where this accuracy is not sufficient, accuracy is most easily improved by adding corrective terms to Newton's equations of motion. Corrective terms account for planetary precession & nutation, GR effects, Coriolis effect, tidal effects, etc.
To give an impression of the magnitudes of such corrections; the second paragraph on the wiki is helpful:
ECI coordinate frames are not truly inertial since the Earth itself is
accelerating as it travels in its orbit about the Sun. In many cases,
it may be assumed that the ECI frame is inertial without adverse
effect. However, when computing the gravitational influence of a third
body such as the Moon on the dynamics of a spacecraft, the
acceleration of the ECI frame must be considered. For example, when
computing the acceleration of an Earth-orbiting spacecraft due to the
gravitational influence of the Moon, the acceleration of the Earth
itself due to the Moon's gravity must be subtracted
The main non-inertial effect to take into account is the acceleration of the ECI frame towards the Moon. If you assume the Earth and Moon are in circular orbit about their barycenter, this acceleration can be estimated;
by Newtonian gravity:
$$
a_{\mathrm{Earth}} = \frac{\mu_{\mathrm{Moon}}}{r_{\mathrm{E-M}}^2} \approx \frac{4902.8\ \mathrm{km}^3\,\mathrm{s}^{-2}}{(384399\ \mathrm{km})^2} = 33.180\, \mu\mathrm{m\,s}^{-2}
$$
from circular motion:
$$
a = \frac{v_{\mathrm{Earth}}^2}{r_{\mathrm{E-M}}} \approx \frac{(1.022\ \mathrm{km\ s}^{-1})^2}{384399\ \mathrm{km}} = 33.180\, \mu\mathrm{m\,s}^{-2}
$$
The same can be said for the Sun. Again, assuming circular orbit:
$$
a_{\mathrm{Earth}} = \frac{\mu_{\mathrm{Sun}}}{r_{\mathrm{S-E}}^2} \approx \frac{132712440018\ \mathrm{km}^3\,\mathrm{s}^{-2}}{(149597871\ \mathrm{km})^2} = 5.9301\, \mathrm{mm\,s}^{-2}
$$