The reason it's not an automatic theorem that "decision is hard implies that counting is hard" is that these two statements use different definitions of "hard".
A decision problem is hard if it's NP-complete under polynomial-time many-one reductions (a.k.a. Karp reductions, a.k.a. polynomial-time mapping reductions).
A counting problem is hard if it's #P-complete under polynomial-time Turing reductions (a.k.a. Cook reductions).
As such, if a decision problem is NP-complete, we know that the corresponding counting problem is NP-hard but that's not the definition of what a hard counting problem is. Being #P-complete seems to be a much stronger statement than just being NP-hard – Toda has shown that #P-complete problems are hard for the entire polynomial hierarchy under randomized reductions so, as a complexity class, #P feels much closer to PSPACE than to NP.
Going in the opposite direction, it's clearly true that, if the counting problem is easy in the sense of being in FP, then the decision problem is in P. After all, if you can efficiently count, you can certainly tell if the answer is nonzero. However, just because the counting version is "not hard" (i.e., not #P-complete) doesn't imply that it's "easy" (i.e., in FP). Ladner's theorem extends to #P so, if FP$\,\neq\,$**#P** then there's an infinite hierarchy of distinct complexity classes between them so our "not-hard" counting problem could be complete for any one of those classes and, therefore, not be "easy" (in FP).
Having said that, I don't think we have any counterexamples to the conjecture that a decision problem being NP-complete means that the counting version is #P-complete. So, it's not a theorem but it is empirically true.