What is the map unit between the genes?
Here's a hint: you're right about 0.01 being the recombinant (crossed-over) double homozygous recessive offspring. I think you're forgetting that each offspring contains 2 gamete haplotypes, and so predicting the percent of offspring with a given genotype involves multiplying the frequencies of the individual gametes that produce that offspring. To calculate the gamete frequencies (and thus the recombination frequency) you need to reverse that multiplication process.
Full answer:
I find it easiest to deal with uniform genotypes first. Of the 4 offspring phenotype categories, only the double homozygous recessive category (short/green) has uniform genotype - llss. If l and s were unlinked, 1/16 of the total offspring would be short/green=llss.
However, many fewer than 1/16 are short/green=llss, so linkage is confirmed and (as you note), ls must represent one of the crossed-over haplotypes. The other crossed-over haplotype is then LS. Thus, the parental (non-crossed-over) haplotypes must be lS and Ls.
Let's call r the recombination distance (in map units) between l and s. Each recombination between l and s produces 2 recombinant gametes: 1 ls and 1 LS.  Thus the fraction of ls gametes - f(ls) - will be (r/2)/100 and the fraction of LS gametes - f(LS) - will also be (r/2)/100.  Since each F1 offspring consists of 2 gametes, the fraction of offspring with an ls/ls genotype will be f(ls) x f(ls) = (r/2)/100 x (r/2)/100 = r^2/40000 = 0.01 (from F1 data).
Solving for r, r^2 = 0.01 x 40000 = 400, so sqrt(r^2) = r = sqrt(400) = 20.
r = 20 map units