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I left the other answer but I just realized from what you said you actually wanted the calculation of the horrible setup you have.

Ok you will need to break the sitation into two the blue shift approach and the red shift recede

So lets map the distances from the emitter and get the time observer 2 passes the particle

Particle: (Simply moves away with time)
distance from emitter = 0.9c * t

Observer 2: (He starts at detector 0.9c away and reduces down)
distance from emitter = 0.9c - 0.8c * t

So they pass each other when time "t" is equal so we have
(0.9c * t) = 0.9c - (0.8c * t)

rearranging
(0.9c * t) + (0.8c * t) = 0.9c

simplifying steps
1.7c * t = 0.9c
t = 0.9c / 1.7c

t = 0.5294 seconds

So if observer 2 passes receiver at time particle leaves emitter they meet 0.5294 sec in journey

Lets just double check that

The particle will move 0.9c for 0.5294 = 0.47646c from the emitter
The observer will move 0.8c for 0.5294 = 0.42352c from detector

Add the two together and hopefully we get our 0.9c the emitter and detector are apart ... 0.47646+0.42352=0.89998 ..... woot no maths fail for me

So observer 2 will see the particle blue shifted for 0.5294 sec then it will pass him and red shift away