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Analogy is a good translation to "liknelse".
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Similarity is another.
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OK, the flow rate / pressure drop relation is logaritmic. So if I meassure all my components separately I can´t assume that the sum of all these flow restriction equals the total flow resistance in my system, when
assembled.
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The relationship bewteen pressure and flow is a square law if that's what you mean by logaritmic. And you CAN pretty much add individual resistances to get a system resistance as long as you assume that the flow is a the same through each resistance.
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Any ideas how I cen determine the flow resistance in two parallell loops, after a Y-splitter?
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Yes. By dividing the flow into two you instead have two half flows with both flows at half speed. Given that pressure drop is a function of the square of flowrate, running a 1/2 flow will give you a 1/4 pressure drop. And the fact that you have two paths agains halfs pressure drop. So you will have 1/8th of the pressure drop of running the flow through 1 WB. This is assuming that flowrate is held constant.
In reality the lower resistance will be compensated by an increase in flow provided by the pump (and assuming that your WB's are the biggest resistance in your loop). This will mean that in practice the flowrate through each WB will be only slightly less in the parallel configuration than if you just had the 1 block, subject to the pump being able to supply the flow.