f1 = log(log*n) and f2 = log*(logn)
log*n means in how may steps logn value will become 1, consider the following example, suppose n=$2^{128}$
log*$2^{128}$: 4 because
log$2^{128}$ -> log$2^{7}$->Log2.80 -> log1.48 -> (approx 1) in 4 steps log*$2^{128}$ value has become 1,
f1 = log(log*n) = log(4) =2
f2 = log*(logn) = log*128 = 3
Hence f1=O(f2), you consider even more larger value than $2^{128}$