proving that if $\{w\$w^R | w \in L\}$ is context-free then $L$ is regular
Here is a proof sketch. The idea is to take a context-free grammar for $L'$ and come up with a regular grammar for $L$.

Start with a grammar for $L'$. Say that a language $M$ is symmetric if it is of the form $\{w \$ w^R : w \in P\}$ for some $P \subseteq \Sigma^*$. Prove that every non-terminal in $L'$ generates a language of one of the forms $w, wM, Mw$, where $w \in \Sigma^*$ and $M$ is symmetric. Consider two cases: the non-terminal generates a word without $\$$, or it generates a word with $\$$. In the first case, it's easy to see that it must generate a unique word. In the second case, every word it generates must contain $\$$, and the words need to be "compatible" in some sense, which implies that it generates a language of the form $wM$ or $Mw$.

For each non-terminal generating $wM$ or $Mw$, we can come up with a non-terminal generating the language $M$. This is because context-free languages are closed under this operation. So we can assume without loss of generality that each non-terminal either generates a unique word or a symmetric language.

Substitute all non-terminals generating unique words. All remaining productions must be of the form $A \to w B w^R$ or $A \to w \$ w^R$. We can come up with an equivalent grammar in which $w$ is always a symbol in the first case and empty in the second, so that all productions are now either $A \to a B a^R$ or $A \to \$$. If we replace these productions with $A \to a B$ and $A \to \epsilon$ then we get a right regular grammar for $L$.

For completeness, let me mention that the converse to the theorem also holds. You can see this by reversing the construction: from a right regular grammar for $L$, construct a context-free grammar for $L'$.