Impedance matching 2MHz clock 2m cable PCB to PCB
Using the rule of thumb of 1/10 wavelength, at 2MHz the 1/10 is 149m,
  which is well below my 8m distance but I would rather tread on the
  safety side


The rule of thumb is broadly correct but you analysis is poor. A clock signal is virtually a square wave and a square wave contains an infinite number of harmonics so, in order to preserve the clock's square (ish) shape you need to ensure that harmonics are considered and not the fundamental frequency of 2 MHz.

Maybe you can get away with the 7th harmonic or maybe it's the 9th, 11th, 13th or fifteenth. You have to decide what you think is about right. If you choose the fifteenth harmonic then the cable distance drops from 149 metres to about 10 metres. Other points: -


The buffer you have chosen is well over-the-top and this is not needed. A decent CMOS buffer will do the job. 
The trace impedance calculator looks fine - model the line with respect to an inner ground/power layer.
Virtually any decent connector will do the job
Differential clock signals are much more preferable because of emissions problems with single-ended clocks