Why is $FOLLOW$ not necessary for $LL(1)$ grammars with no $\epsilon$ transitions?
That's correct for $LL(1)$: if there are no $\epsilon$ productions, the $LL$ parser-generation algorithm will never consult $FOLLOW$, because it only does that if it finds $\epsilon$ in the $FIRST$ set for the first non-terminal in the right-hand side of a production. (So it might not need the $FOLLOW$ sets even if there are some $\epsilon$ productions, provided that none of those productions occur at the beginning of a right-hand side.) 

The observation doesn't generalise well to other values of $k$. You'll need $FOLLOW_k(\alpha)$ if any production can derive a string whose length is less than $k$.