For the 64-bit multiplication, you're adding 22 terms, one for each base-8 digit. (Think of grade-school multiplication.) Each term needs to be trivial to compute, so you can do a shift and/or a negation to get the term, but you can't do another addition.

The point is that ×3 is precomputed once, so you can then simply feed it into any of the 22 terms as needed. You can't do ×2 and ×1 into a term to get ×3 because every term would need another adder. In other words, you want one circuit to compute ×3 rather than 22 circuits.

For your ×6 question, this value is computed by putting negative ×2 into the term and conceptually adding one more to the next digit to get ×8. You can't right-shift this ×8 value since it's part of a completely different term.

Hopefully this makes sense. There are a lot of digits and sums flying around.