[proofplan]
Fix $\xi$ and restrict $b$ to the line segment from $x$ to $y$, obtaining a one-variable smooth function of the segment parameter. The one-variable Taylor formula with integral remainder gives an expansion in derivatives of this restricted function. The chain rule and multinomial formula identify those derivatives with the multi-index derivatives $\partial_x^\alpha b$ and produce exactly the coefficients in the statement. Finally, the symbol estimates for $b$ on the compact segment show that each integral remainder term has the same $\xi$-symbol order $m$.
[/proofplan]
[step:Restrict the symbol to the line segment from $x$ to $y$]
Fix $\xi \in \mathbb{R}^n$. Define the segment map
\begin{align*}
\gamma: [0,1] \to U,\qquad \gamma(t):=x+t(y-x).
\end{align*}
This map is well-defined because the hypothesis gives $[x,y]\subset U$. Define
\begin{align*}
f_\xi: [0,1] \to \mathbb{C},\qquad f_\xi(t):=b(\gamma(t),\xi).
\end{align*}
Since $b$ is smooth in the $x$ variable on $U$ and $\gamma$ is smooth, $f_\xi$ is $C^\infty$ on $[0,1]$.
Applying the one-variable Taylor formula with integral remainder to $f_\xi$ at $0$ and evaluating at $1$ gives
\begin{align*}
f_\xi(1)=\sum_{k=0}^{N-1}\frac{f_\xi^{(k)}(0)}{k!}+\frac{1}{(N-1)!}\int_0^1(1-t)^{N-1}f_\xi^{(N)}(t)\,d\mathcal{L}^1(t).
\end{align*}
Because $f_\xi(1)=b(y,\xi)$ and $f_\xi(0)=b(x,\xi)$, it remains to rewrite the derivatives $f_\xi^{(k)}$ in multi-index form.
[guided]
The reason for introducing $f_\xi$ is that Taylor expansion in the spatial variable $x$ can be reduced to Taylor expansion in one real variable along the line segment from $x$ to $y$. We fix $\xi \in \mathbb{R}^n$ and define
\begin{align*}
\gamma: [0,1] \to U,\qquad \gamma(t):=x+t(y-x).
\end{align*}
The map $\gamma$ is legitimate because the theorem assumes the whole segment $[x,y]$ lies in $U$. Now define
\begin{align*}
f_\xi: [0,1] \to \mathbb{C},\qquad f_\xi(t):=b(\gamma(t),\xi).
\end{align*}
The symbol $b$ is smooth in the spatial variable, and $\gamma$ is a smooth affine map, so the composition $f_\xi$ is a smooth one-variable function.
The one-variable Taylor formula with integral remainder, applied at $t=0$ and evaluated at $t=1$, gives
\begin{align*}
f_\xi(1)=\sum_{k=0}^{N-1}\frac{f_\xi^{(k)}(0)}{k!}+\frac{1}{(N-1)!}\int_0^1(1-t)^{N-1}f_\xi^{(N)}(t)\,d\mathcal{L}^1(t).
\end{align*}
Here the measure in the integral is one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal{L}^1$. Since $\gamma(0)=x$ and $\gamma(1)=y$, this formula is already a Taylor expansion of $b(y,\xi)$ around $x$; the remaining work is to express $f_\xi^{(k)}$ through the partial derivatives of $b$.
[/guided]
[/step]
[step:Compute the line derivatives by the chain rule and multinomial formula]
Let $h:=y-x \in \mathbb{R}^n$. For each integer $k\geq 0$ and each $t\in[0,1]$, repeated application of the chain rule gives
\begin{align*}
f_\xi^{(k)}(t)=\left(\sum_{j=1}^n h_j\partial_{x_j}\right)^k b(x+th,\xi).
\end{align*}
Expanding the $k$-th power by the multinomial formula,
\begin{align*}
f_\xi^{(k)}(t)=\sum_{|\alpha|=k}\frac{k!}{\alpha!}h^\alpha\partial_x^\alpha b(x+th,\xi).
\end{align*}
In particular,
\begin{align*}
\frac{f_\xi^{(k)}(0)}{k!}=\sum_{|\alpha|=k}\frac{(y-x)^\alpha}{\alpha!}\partial_x^\alpha b(x,\xi).
\end{align*}
For $k=N$,
\begin{align*}
\frac{1}{(N-1)!}f_\xi^{(N)}(t)=\sum_{|\alpha|=N}\frac{N(y-x)^\alpha}{\alpha!}\partial_x^\alpha b(x+t(y-x),\xi),
\end{align*}
because
\begin{align*}
\frac{N!}{(N-1)!}=N.
\end{align*}
[/step]
[step:Substitute the computed derivatives into Taylor's formula]
Substituting the identities from the previous step into the one-variable Taylor formula yields
\begin{align*}
b(y,\xi)=\sum_{k=0}^{N-1}\sum_{|\alpha|=k}\frac{(y-x)^\alpha}{\alpha!}\partial_x^\alpha b(x,\xi)+\sum_{|\alpha|=N}\frac{N(y-x)^\alpha}{\alpha!}\int_0^1(1-t)^{N-1}\partial_x^\alpha b(x+t(y-x),\xi)\,d\mathcal{L}^1(t).
\end{align*}
Since the condition $0\leq k\leq N-1$ is equivalent to $|\alpha|<N$, the first double sum is
\begin{align*}
\sum_{|\alpha|<N}\frac{(y-x)^\alpha}{\alpha!}\partial_x^\alpha b(x,\xi).
\end{align*}
This proves the displayed expansion.
[/step]
[step:Estimate each remainder term as a symbol of order $m$]
Fix a multi-index $\alpha$ with $|\alpha|=N$. Define the compact segment
\begin{align*}
C_{x,y}:=\{x+t(y-x):t\in[0,1]\}\subset U.
\end{align*}
For a multi-index $\beta \in \mathbb{N}_0^n$, differentiating $R_\alpha^{x,y}$ in $\xi$ under the integral gives
\begin{align*}
\partial_\xi^\beta R_\alpha^{x,y}(\xi)=\frac{N(y-x)^\alpha}{\alpha!}\int_0^1(1-t)^{N-1}\partial_\xi^\beta\partial_x^\alpha b(x+t(y-x),\xi)\,d\mathcal{L}^1(t).
\end{align*}
This differentiation under the integral is justified because the integrand is smooth in $\xi$ and continuous in $t$ on the compact interval $[0,1]$.
By the defining estimates for $b\in S^m_{1,0}(U\times\mathbb{R}^n)$, there exists a constant $M_{\alpha,\beta,x,y}>0$ such that
\begin{align*}
|\partial_\xi^\beta\partial_x^\alpha b(z,\xi)|\leq M_{\alpha,\beta,x,y}(1+|\xi|)^{m-|\beta|}
\end{align*}
for every $z\in C_{x,y}$ and every $\xi\in\mathbb{R}^n$. Therefore
\begin{align*}
|\partial_\xi^\beta R_\alpha^{x,y}(\xi)|\leq \frac{N|(y-x)^\alpha|}{\alpha!}M_{\alpha,\beta,x,y}(1+|\xi|)^{m-|\beta|}\int_0^1(1-t)^{N-1}\,d\mathcal{L}^1(t).
\end{align*}
The one-dimensional integral is
\begin{align*}
\int_0^1(1-t)^{N-1}\,d\mathcal{L}^1(t)=\frac{1}{N}.
\end{align*}
Hence
\begin{align*}
|\partial_\xi^\beta R_\alpha^{x,y}(\xi)|\leq \frac{|(y-x)^\alpha|}{\alpha!}M_{\alpha,\beta,x,y}(1+|\xi|)^{m-|\beta|}.
\end{align*}
Setting
\begin{align*}
C_{\alpha,\beta,x,y}:=1+\frac{|(y-x)^\alpha|}{\alpha!}M_{\alpha,\beta,x,y}
\end{align*}
gives $C_{\alpha,\beta,x,y}>0$ and proves that $R_\alpha^{x,y}$ is a symbol of order $m$ in $\xi$. Since there are only finitely many multi-indices $\alpha$ with $|\alpha|=N$, the full remainder is a finite sum of terms of $\xi$-symbol order $m$.
[/step]
