[guided]The next task is to identify exactly which terms can still contribute after the order $m+k$ cancellation. There are two possible sources of degree $m+k-1$: the $\alpha=0$ part using one subprincipal homogeneous term, and the $|\alpha|=1$ part using the two leading homogeneous terms.
First consider $\alpha=0$. In $q_{PA}$, the degree $m+k-1$ part coming from $\alpha=0$ is
\begin{align*}
p\,a_{k-1}+p_{m-1}\,a.
\end{align*}
In $q_{AP}$, the corresponding expression is
\begin{align*}
a\,p_{m-1}+a_{k-1}\,p.
\end{align*}
These expressions cancel in the difference because the symbols are scalar-valued functions:
\begin{align*}
p\,a_{k-1}=a_{k-1}\,p
\end{align*}
and
\begin{align*}
p_{m-1}\,a=a\,p_{m-1}.
\end{align*}
This cancellation is precisely why the theorem assumes scalar principal symbols; for matrix-valued leading symbols, an order $m+k$ matrix commutator can remain.
Now consider the terms with $|\alpha|=1$. For each $j\in\{1,\dots,n\}$, let $e_j\in\mathbb{N}_0^n$ denote the multi-index with $1$ in the $j$th position and $0$ elsewhere. The $e_j$ term in the composition formula for $PA$ is
\begin{align*}
\partial_{\xi_j}q_P\,D_{x_j}q_A.
\end{align*}
Taking the leading homogeneous part of this expression gives
\begin{align*}
\partial_{\xi_j}p\,D_{x_j}a,
\end{align*}
because $\partial_{\xi_j}p$ has degree $m-1$ in $\xi$ and $D_{x_j}a$ has degree $k$ in $\xi$. Thus the degree is $m+k-1$. Summing over $j$ gives the degree $m+k-1$ contribution
\begin{align*}
\sum_{j=1}^n \partial_{\xi_j}p\,D_{x_j}a.
\end{align*}
The same argument applied to $AP$ gives
\begin{align*}
\sum_{j=1}^n \partial_{\xi_j}a\,D_{x_j}p.
\end{align*}
No term with $|\alpha|\geq 2$ can contribute to degree $m+k-1$. Each derivative $\partial_{\xi_j}$ lowers the homogeneous degree by one, while $D_{x_j}$ differentiates only in the base variable and does not change the homogeneous degree in $\xi$. Therefore a term with $|\alpha|\geq 2$ has degree at most $m+k-2$ when applied to the two leading homogeneous symbols, and lower-order homogeneous pieces only decrease the degree further.
Consequently the homogeneous component of degree $m+k-1$ in the commutator symbol is
\begin{align*}
\sum_{j=1}^n\left(\partial_{\xi_j}p\,D_{x_j}a-\partial_{\xi_j}a\,D_{x_j}p\right).
\end{align*}
Now insert the convention $D_{x_j}=\frac{1}{i}\partial_{x_j}$. This gives
\begin{align*}
\frac{1}{i}\sum_{j=1}^n\left(\partial_{\xi_j}p\,\partial_{x_j}a-\partial_{\xi_j}a\,\partial_{x_j}p\right).
\end{align*}
Since the symbols are scalar-valued, multiplication commutes, so
\begin{align*}
\partial_{\xi_j}a\,\partial_{x_j}p=\partial_{x_j}p\,\partial_{\xi_j}a.
\end{align*}
Thus the expression becomes
\begin{align*}
\frac{1}{i}\sum_{j=1}^n\left(\partial_{\xi_j}p\,\partial_{x_j}a-\partial_{x_j}p\,\partial_{\xi_j}a\right).
\end{align*}
By the stated coordinate formula for the Poisson bracket on $T^*U$, this is exactly
\begin{align*}
\frac{1}{i}\{p,a\}.
\end{align*}[/guided]