Here's a possible reduction for $A_{tm} \le S_{tm}$.
Map input $\langle M,w\rangle$ to $\langle M', 0w1\rangle$ such that
$M'$ on input $x$:
0. if $x$ can be written as $1x'0$ for some $x'$, accept. Otherwise,
- if $x$ can be written as $0x'1$ for some $x'$, run $M$ on $x'$ and output the same.
- Otherwise, accept (or reject, it doesn't matter).
proof.
If $\langle M,w\rangle \in A_{TM}$ then $M$ accepts $w$. Observe that $M'$ accepts both $0w1$ (rule 1) and $1w^r0$ (rule 0), thus $\langle M', 0w1\rangle \in S_{TM}$.
On the other hand, if $\langle M,w\rangle \notin A_{TM}$, then also $M'$ doesn't accept $0w1$, and $\langle M', 0w1\rangle \notin S_{TM}$.
Observe that both $A_{TM}$ and $S_{TM}$ are $RE$ languages (which are not in $R$). You can also prove a reduction $S_{TM} \le A_{TM}$, by mapping $\langle M,w\rangle$ to $\langle M',w\rangle$, with $M'$ that runs $M$ on $w$ and then on $w'$. The proof is straightforward.