$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\const}{\operatorname{const}}$
As $S(x,t)$ satisfies equation equation (5.2.1) \begin{equation*} P_0(x,\nabla S(x))=0 \end{equation*} (we again consider $x_0=t$ as part of $x$) consider the next term in the right-hand expression (5.1.6); it has factor $h$ and we equalize it to $0$ ignoring so far all other terms: \begin{equation} \bigl(-i\sum_j P_0^{(j)} (x, \nabla S(x))\partial_j +Q(x)\bigr)A (x)=0 \label{eq-5.3.1} \end{equation} where \begin{equation} Q(x)=-\frac{i}{2}\sum _{j,k} P_0^{(jk)}(x,\nabla S(x)) S_{x_jx_k} (x) +P_1(x,\nabla S(x)) \label{eq-5.3.2} \end{equation}
Equation (\ref{eq-5.3.1}) to amplitude $A$ is a linear ODE along trajectory (5.2.5). Indeed, \begin{equation*} \sum_j P_0^{(j)} (x, \nabla S(x))\partial_j = \frac{d\ }{d\tau}; \end{equation*} so (\ref{eq-5.3.1}) becomes \begin{equation} \bigl(\frac{d\ }{d\tau} +iQ(x)\bigr)A (x)=0 \label{eq-5.3.3} \end{equation}
Let us look at $Q(x)$. Observe that according to Liouville theorem along trajectories \begin{multline} \frac{d\ }{d\tau} \ln J(x,t) = \sum_j \partial_{x_j} \bigl(P_0^{(j)}(x,\nabla S(x))\bigr)=\\ \sum_j \partial_{x_j} P_{0(j)}^{(j)}(x,\nabla S(x)) + \sum_{j,k} P_{0}^{(jk)}(x,\nabla S(x))S_{x_jx_k} \label{eq-5.3.4} \end{multline} where $J$ is a volume element. Recall that $P_{0(j)}$ is calculated as if $\nabla S$ does not depend on $x$. Then (\ref{eq-5.3.3}) becomes \begin{equation} \bigl(\frac{d\ }{d\tau} +i P^s(x,\nabla S)\bigr)A (x)J^{1/2}=0 \label{eq-5.3.5} \end{equation} with \begin{equation} P^s := P_1 -\frac{i}{2} \sum_j P_{0(j)}^{(j)}. \label{eq-5.3.6} \end{equation}
In particular \begin{equation} \bigl(\frac{d\ }{d\tau} +2\Re i P^s(x,\nabla S)\bigr)|A (x)|^2J=0. \label{eq-5.3.7} \end{equation}
Remark 1. In fact $P^s$ rather than $P_1$ is a "correct" expression. All symbols $P_l$ with $l\ge 1$ depend on he method of quantization. We used quantization in which $p=-ih\nabla$ acts first and then $q$ (aka $x$). But this violates lot of things. Correct quantization was invented by Hermann Weyl and is called Weyl quantization (or symmetric quantization) and in this quantization $P^s$ is the next symbol!
Usually in Microlocal analysis $P_0$ is called principal symbol and $P^s$ is called subprincipal symbol. For Hermitian and self-adjoint operators all Weyl symbols are real (and v.v.) so in this case $\frac{d\ }{d\tau} |A|^2J=0$ and $|A|^2$ is a density.
In fact there are many terms in the right-hand expression (5.1.6). Because of this we need amplitude $A(x)$ in the form \begin{equation} A(x)\sim \sum_{n\ge 0} A_0(x)h^n \label{eq-5.3.8} \end{equation} and $A_0$ rather than $A$ satisfies all above equations while $A_n$ with $n\ge 1$ takes care of the term with $h^{n+1}$: \begin{equation} \bigl(\frac{d\ }{d\tau} +iQ(x)\bigr)A _n(x)=\sum_{l\le n-1} \mathcal{L}_{n-l} A_l \label{eq-5.3.9} \end{equation} where $\mathcal{L}_{n-l}$ are partial differential operators of order $n-l+1$.