Why does the NP completeness of the Hartree-Fock method not lead to difficulty in practical calculation?
There is plenty of precedent for problems that are NP-hard (in worst-case complexity) but can often be done in practice on many instances that arise in practice (e.g., because their average-case complexity is low, or for other reasons).

NP-hardness relates to the worst-case complexity of a problem.  However, worst-case hardness isn't always representative of typical-case hardness.  Sometimes typical instances are much easier than the worst-case instances.  Therefore, NP-hardness results aren't always an indication that the problem can't be solved acceptably well in practice.

As Yuval Filmus explained so well in a comment:


  I'm not sure there is any better answer than the fact that the algorithms work on real-world instances. NP-hardness is a worst-case concept, and sometimes the worst case is simply irrelevant. It is a challenging problem to characterize which properties of real-world instances render the problem solvable. Smooth analysis is one answer, and there might be others in the non-rigorous literature.


I recommend you take a look at Dealing with intractability: NP-complete problems for an overview of common methods for dealing with NP-hardness: i.e., for figuring out ways to solve a problem good enough for practical purposes, when we know it is NP-hard.