Let $G$ be a finite group acting on itself by the action $g \ast x := xg^{-1}$. Then this corresponds to an representation $\rho : G \to GL(L^2(G))$, where $L^2(G)$ denotes the space of function on $G$ (this is called the right regular representation). This is given for $f \in L^2(G)$ by $$ (\rho(g)f)x := f(xg) $$ i.e. the arguments of $f$ are "translated" by $g$. That this is a representation could be found here. Now we have $L^2(G) \cong F_G$, where $F_G$ denotes the free vector space on $G$, where the element $$ \sum_{x \in G} a_x g \in F_G $$ corresponds to the function $f(x) = a_x$. With this $$ g \ast \left( \sum_{x \in G} a_x g \right) = \sum_{x\in G} a_x \cdot xg^{-1} = \sum_{x \in G} a_{xg} \cdot x $$ which corresponds to the function $f(xg)$, hence $g \ast f = \rho(g)f$.
Now I have a question on the character of the right regular representation, we have $$ \chi_{reg} = \sum_{i}^d d_i \chi_i $$ where $d_i = \dim V_i = \chi_i(1)$ and $\chi_i$ are the characters for the irreducible representations $V_i$ of $G$. This is stated in these lecture notes, and in the proof it is just mentioned that this follows from $$ L^2(G) \cong \bigoplus_i R_i, \quad R_i \cong d_i \cdot V_i \quad \mbox{(i.e. $V_i\oplus\cdots\oplus V_i$ with $d_i = \dim V_i$ copies)} $$ I do not see the connection between the direct sum of $L^2(G)$ and the linear combination of $\chi_{reg}$? Can someone please explain?