Why is $A_{TM}$ reducible to $HALT_{TM}$?
Step one, don't try to argue why this TM can not work: under the assumption it does work. But then, since everything else $S$ does is possible (you have to accept that separately), the existence of $R$ certainly is "the problem".

As for understanding, the proof, there are two facts you have to check (under the assumption that $R$ exists):


$S$ is a Turing machine (i.e. its function is computable)
$S$ decides $A_{TM}$.


Arguably, 1) is clear; simulating other TMs given their indices/encodings is something TMs can do (thanks to the existence of a universal TM, which you should have seen a proof of already) and $S$ does little more. This is as close to a proof as you'll get with this form of "definition" of $S$; it's more of an idea, really.

For the second, note that $S$ always halts because $R$ always halts and

$\qquad\begin{align*}
  S\langle M,w \rangle) = 1 &amp;\iff R(M,W) = 1 \land M(w) = 1 \\
             &amp;\iff M(w)\downarrow \land M(w) = 1 \\
             &amp;\iff w \in L_M \\
             &amp;\iff \langle M,w \rangle \in A_{TM} \;.
\end{align*}$

By definition, that means that $S$ decides $A_{TM}$.

Similar arguments usually works for this kind of proof; check our reference questions and other questions tagged computability+reductions.