Let $G$ be a compact group. Then there is an isomorphism $L^2(G)\simeq \bigoplus_{\tau\in \hat{G}} V_{\tau}\otimes V_{\tau^*}$ which intertwines the conjugation action of $G\times G$ on $L^2(G)$ $$C(g_1, g_2)f (x)=f(g_1^{-1}xg_2)=L(g_1)R(g_2)f (x)$$ and the $G\times G$ action on sum of $V_{\tau}\otimes V_{\tau^*}$ spaces defined by $$A(g_1, g_2)(v\otimes f)= \tau(g_1)v\otimes \tau^*(g_2)f$$ for each pure tensor $v\otimes f\in V_{\tau}\otimes V_{\tau^*}$.
(Here $\tau^*$ denotes the dual of the representation $\tau$.)
Upon restriction of this intertwiner to $e\times G$ I expect to get an isomorphism $$R\simeq \bigoplus_{\tau\in \hat{G}}1_{\tau}\otimes \tau^*$$ for the right regular representation.
However, I have seen this isomorphism being written differently as $$R\simeq \bigoplus_{\tau\in \hat{G}}\tau\otimes 1_{\tau^*}.$$
Can someone, please, clarify which version is correct and why?
In your notation, you get the second isomorphism by restricting to $G\times \{e\}$, the left regular representation, and not the right regular one. But the two representations are isomorphic via the map $x\to x^{-1}$; therefore, there is no conflict.
Note also that while your notation makes the first isomorphism more immediately believable, there no sense in which it is 'canonically' better; just note that switching $g_1$ and $g_2$ compatibly in the definitions of both $C$ and $A$, you get the other isomorphism for the right regular representation.