Mapping reductions for dummies
The idea of the reduction you give in your question is the following:

I) If $\langle M,w\rangle \in A_{tm}$, then $L(M')=\Sigma^*$.
II) If $\langle M,w\rangle \notin A_{tm}$, then $L(M')=\{10\}$.

thus, for any $w$, $\langle M,w\rangle \in A_{tm}$ iff $\langle M',w\rangle \in S_{tm}$.
In case (I), for any $w\in L(M')$ also $w^R\in L(M')$ hence $\langle M',w \rangle S_{TM}$ for any $w$. For case (II), there is no $w\in L(M')$ such that $w^R\in L(M')$ as well, thus for any $w$, $\langle M',w \rangle S_{TM}$.

Now for your question about accepting "10" no matter what: Assume we avoid this step. Then, in case (II), $L(M')=\emptyset$, hence $\langle M',w\rangle \in S_{tm}$ for any $w$, since the condition on $w,w^R$ vacuously holds. Now the mapping cannot map inputs  $\langle M,w\rangle \notin A_{tm}$ to inputs outside $S_{TM}$ (since none exists).

The value "10" is arbitrary. Any other non-palindromic string will do.