Let $G = GL(2,\textbf{R})^+$ be the $2 \times 2$ invertible matrices with positive determinant and given some discontinuous subgroup $\Gamma$, let $V = L^2(\Gamma \backslash G, \chi)$, the space of measurable complex-valued functions $f$ on $G$ satisfying $$ f(\gamma g u) = \chi(\gamma)F(g), \quad \gamma \in \Gamma, u \in Z', g \in G. $$ (where $Z'$ is subgroup of scalar matrices with positive diagonal) and that are square-integrable with respect to the Haar measure on $G/Z' \cong SL(2,\textbf{R})$. The space $V$ is often denoted $V = L^2(\Gamma \backslash G / Z', \chi) = L^2(\Gamma \backslash SL(2,\textbf{R}), \chi)$.
Consider the right regular representation $R: G \to End(V)$, given by $$R(g)f(x) = f(xg).$$ To verify that this is a representation, one needs to check that $R(gg')f = R(g)R(g')f$. This is easy enough to do directly: $$R(gg')f(x) = f(xgg') = (R(g')f)(xg) = R(g)(R(g')f)(x).$$
However, a mistake that I kept making was when I tried to go backwards instead and kept applying $R(g')$ before $R(g)$, like for function composition, from which I got $$R(g)R(g')f(x) = R(g)f(xg') = f(xg'g) \neq R(gg')f(x).$$
What is the reason that "applying R(g') first" is wrong?