NP-Complete Hard-on-Average Problems
If there exist some NP language L that's hard on the average on some distribution $\mu$, then by applying the standard reduction $f$ from L to 3SAT you get that 3SAT is hard on $f(\mu)$.

Your understanding is wrong. By saying that the inverting task  for a function f is NP hard, one means that there is a polynomial-time algorithm A such that A can solve 3SAT on every input given black-box access to an inverter I that inverts f on a non-negligible fraction of the inputs.