Let $\rho$ be a field automorphism of $K$. Let $\sigma:K\to\mathbb C$ be a field embedding into the complex numbers $\mathbb C$. Let $x\in K$.
$\textbf{Q:}$ What is the meaning of $x^{\rho\circ\sigma}$? My first guess is that it should be $x^{\sigma\circ \rho}=\sigma(\rho(x))$. I could not understand this notation there is no guarantee that $\sigma(x)\in K$ for $x\in K$. In the notation $x^{\rho\circ\sigma}=\rho(\sigma(x))$. So $\sigma(x)$ might be even outside the domain of definition of $\rho$. This is related to the fact that if $\rho$ can be extended to an automorphism of $C$ as complex conjugation, then $\rho$ will induce a valued field automorphism by pulling back the absolute value on $C$.
This is Taylor Frohlich Algebraic number theory Pg 64's 2.10.b)
The notation $x^\sigma$ stands for $\sigma^{-1}(x)$: it is a right action, so this is the way to ensure that $(x^\sigma)^\tau = x^{\sigma\tau}$ holds. So in your case, $x^{\rho\sigma}$ is $(\rho\sigma)^{-1}(x) = \sigma^{-1}\rho^{-1}(x)$, which is perfectly fine, since $\rho^{-1}$ preserves $K$.