Well that solution looks over-complicated to me but it has nothing much to do with the maths.

The first step in solving this problem has to be your understanding of what they are trying to do. I can see what they are trying to do, but can you? If you can’t then you will not solve your problem because the characteristics of your transducers will be different to theirs and you will have no hope of making the system work.

Spell out in simple English what you / they are trying to achieve and we will take it from there.

What is the end result of this project?  Perhaps there is something off the shelf that can be used.

Allright. They're trying to combine the slow characteristics of the inclinometer and the fast characteristic of the gyro to determine a motion's(velocity) orientation.

Since inclinometers work "best" with slow motions, and gyros with fast, they filter these signals to fit, or supplement, with each other.

As far as I can understand we need a HP filter for the inclinometer and a LP filter for the gyro. The question is now how to design these filters?

Swap your HP/LP requirements, else you're going to get no useful signals out...

Fatpo says it all(except for the LP-HR typo).  The other thing the filter is doing is an approximate compensation for the inclinometer non-linearity.

For an introduction to filters could I suggest the book "The Art of Electronics."  And then for digital filters "DSP Guide."

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.