Reducing A(TM) to some decidable problem
$A(TM)$ is undecidable because a Turing machine can fail to halt on its input. $A(DFA)$ is decidable since a DFA must always halt on any input. We can always decide whether a DFA accepts some string just by running the DFA on the string and seeing whether the DFA ends up in an accepting state. Not so with a Turing machine.

It is extraordinarily unlikely that there is a valid reduction of $A(TM)$ to $A(DFA)$; that would be a bit like effectively demonstrating that triangles have actually always had 3.5 sides, not 3. If you've got such a proof you should buy as many futures in mops as you can, since there will be a lot of brain matter to clean up after all the computer scientists' and mathematicians' heads explode.

I suspect the real question may really be more about whether proofs of the form "A reduces to B. B is decidable. Therefore, A is decidable." are valid. The answer is that, as you probably expect, of course they're valid: if you ever find yourself needing to prove a problem is decidable, this is the gold standard way to do it. If you've never seen a proof like this, per se, the reasons could be either that the default assumption is that problems are decidable until proven otherwise, or because this proof method is tantamount to just designing an algorithm to solve the problem (in which case, it might not appear to be a proof of anything so much as a description of an algorithm).

To give an example: consider Bubble Sort. This can be viewed as a proof that the Sorting problem is decidable, since it reduces $Sort$ to $\{GreaterThanOrEqualTo, Swap, Loop\}$, which we know are decidable. All right, there's some hand-waving there, but you get the idea.