Ok, you are happy with the basic concept, so we just have to interpret the maths.
You will appreciate that these equations are written in terms of the Laplace transform variable "s", but for our purposes we might just as well replace the "s" with "jw" where "j" is the square root of minus one and "w" is supposed to be a lower case omega, the angular frequency (=2*PI*f).
Section A on page 393 assumes that the low pass filtering of the inclinometer is negligible. That is rather silly since that is the whole problem they are trying to fix! Let’s fast forward to section B then. Equation 7 has the dynamics of the inclinometer included by means of the 1/(0.53s + 1) term. We can translate this as a single pole low pass roll-off at 0.3Hz. You would need to measure the frequency response of your inclinometer and substitute its value here.
You may not have realised it, but there is no deterministic path between equations 7 and 8. Equation 7 has an infinite number of possible solutions and the authors have merely picked one which easily suited their ideas.
Equation 8 is the filter you want to place in the inclinometer path and your problem is apparently that you don’t understand what it means. Well replace the s by j*2*PI*f and it should be evident (provided you understand complex numbers!). The (1.06s + 1) on top is a "zero". On the typical frequency response plots with dB amplitude scaling and log frequency scaling this would appear as a horizontal line until the zero, at which point the line would head upwards at a rate to 20dB/decade of frequency. The (0.53s + 1) underneath is a pole as discussed earlier. The pole is a 20dB/decade drop with frequency starting at 0.3Hz. The zero is a 20dB/decade increase with frequency starting at 0.15Hz. The zero starts first then the pole cancels it, so the magnitude response is 0dB at very low frequencies. Around 0.15Hz it is rising and it levels out around 0.30Hz. The resultant gain above 0.3Hz is x2 (6dB). Clearly this is not a passive filter.
One possible implementation, and again there are many ways of producing this result, is to use a pair of equal valued resistors as a potential divider with a capacitor across the top resistor. The output of the potential divider feeds a x2 buffer amplifier. Run this up on SPICE to get the idea if it is too difficult to understand.
