Here is my two pennies worth:
https://docs.google.com/drawings/d/1_7DDbiSNEaioELLo3h8gCKkSIJmK6d7MawU1ONUOJds/edit?usp=sharing
This is based upon the fact that distance "a" reduces to distance "b" after balloon pop.
I'm assuming "a" is at least 5m and "b" will be less than 3m.
A set of electrical contacts on long, self closing arms that touch when the parachute starts to open. The upper open end of the arms would be tethered to the top inside centre of the chute, and the lower end attached to the platform below (pretty much where length "b" arrow is).
The overall length of the open arms (which would be 2 x 1m = 2m) and the two bits of tether would be equal to distance "a".
During ascent, the weight of the platform and the lift of the balloon keep the arms open (and so the electrical circuit open). In the event of turbulence, the distance "c" would have to reduce by 2m before the circuit would close, hopefully dampening temporary reductions of "c" < "a" and stopping any premature firing. (I know the un-deployed chute will go squiggly as well - couldn't be bothered to draw it.)
After the balloon bursts, the arms will have freedom to close completely and so close the backup firing circuit .
Have the arms below the closed chute material (where the squiggly bit is in measurement "c") so they can't become tangled in the chute.
The firing system could also be a tube like you show in your "pin pull" system, but with the cylinder being upright, about 2m long and the electrical contact being made as the spring pulls the piston all the way in to the tube after about 2m of travel.