[guided]The input space is allowed to have negative Sobolev order, so we cannot bound the local $L^2$ norm of $u$ by $\|u\|_{H^{-b}}$. Instead, we test the output against an $L^2$ function and move the required input regularity onto the kernel through derivatives in the $y$ variable.
Fix $\gamma\in\mathbb{N}_0^n$ with $|\gamma|\le a$ and fix a test function $\varphi\in C_c^\infty(\mathbb{R}^n)$. Since $K_{\chi,\psi}$ is smooth and supported in $X\times Y$, the integrand below is smooth and compactly supported in $X\times Y$. Fubini's theorem therefore applies and gives
\begin{align*}
\int_{\mathbb{R}^n}D_x^\gamma(\chi R_h\psi u)(x)\overline{\varphi(x)}\,d\mathcal{L}^n(x)=\int_{\mathbb{R}^n}u(y)\overline{G_{\gamma,\varphi}(y;h)}\,d\mathcal{L}^n(y),
\end{align*}
where the function $G_{\gamma,\varphi}(\cdot;h):\mathbb{R}^n\to\mathbb{C}$ is defined by
\begin{align*}
G_{\gamma,\varphi}(y;h)=\int_X \overline{D_x^\gamma K_{\chi,\psi}(x,y;h)}\varphi(x)\,d\mathcal{L}^n(x).
\end{align*}
This function is smooth because the kernel is smooth, and it is supported in $Y$ because $K_{\chi,\psi}(x,y;h)=0$ for $y\notin Y$.
To pair $u\in H^{-b}(\mathbb{R}^n)$ with $G_{\gamma,\varphi}$, we must prove that $G_{\gamma,\varphi}\in H^b(\mathbb{R}^n)$ with a rapidly small norm. Let $\delta\in\mathbb{N}_0^n$ satisfy $|\delta|\le b$. Differentiating under the integral sign in $y$ is justified by smoothness and compact support, and yields
\begin{align*}
D_y^\delta G_{\gamma,\varphi}(y;h)=\int_X \overline{D_y^\delta D_x^\gamma K_{\chi,\psi}(x,y;h)}\varphi(x)\,d\mathcal{L}^n(x).
\end{align*}
Apply the Cauchy-Schwarz inequality on the measure space $(X,\mathcal{B}(X),\mathcal{L}^n)$ to the two functions $x\mapsto D_y^\delta D_x^\gamma K_{\chi,\psi}(x,y;h)$ and $x\mapsto\varphi(x)$. The rapid derivative bound for the localized kernel gives
\begin{align*}
|D_y^\delta G_{\gamma,\varphi}(y;h)|\le \mathcal{L}^n(X)^{1/2}C_{\gamma,\delta,M,\chi,\psi}h^M\|\varphi\|_{L^2(\mathbb{R}^n)}.
\end{align*}
Because $D_y^\delta G_{\gamma,\varphi}$ is supported in $Y$, integrating this pointwise estimate over $Y$ with respect to $\mathcal{L}^n$ gives
\begin{align*}
\|D_y^\delta G_{\gamma,\varphi}(\cdot;h)\|_{L^2(\mathbb{R}^n)}\le \mathcal{L}^n(X)^{1/2}\mathcal{L}^n(Y)^{1/2}C_{\gamma,\delta,M,\chi,\psi}h^M\|\varphi\|_{L^2(\mathbb{R}^n)}.
\end{align*}
Summing this estimate over all $\delta$ with $|\delta|\le b$ proves
\begin{align*}
\|G_{\gamma,\varphi}(\cdot;h)\|_{H^b(\mathbb{R}^n)}\le C_{\gamma,b,M,\chi,\psi}h^M\|\varphi\|_{L^2(\mathbb{R}^n)}.
\end{align*}
Now the definition of $H^{-b}(\mathbb{R}^n)$ as the dual of $H^b(\mathbb{R}^n)$ gives
\begin{align*}
\left|\int_{\mathbb{R}^n}u(y)\overline{G_{\gamma,\varphi}(y;h)}\,d\mathcal{L}^n(y)\right|\le \|u\|_{H^{-b}(\mathbb{R}^n)}\|G_{\gamma,\varphi}(\cdot;h)\|_{H^b(\mathbb{R}^n)}.
\end{align*}
Combining the preceding estimates, we obtain
\begin{align*}
\left|\int_{\mathbb{R}^n}D_x^\gamma(\chi R_h\psi u)(x)\overline{\varphi(x)}\,d\mathcal{L}^n(x)\right|\le C_{\gamma,b,M,\chi,\psi}h^M\|u\|_{H^{-b}(\mathbb{R}^n)}\|\varphi\|_{L^2(\mathbb{R}^n)}.
\end{align*}
Taking the supremum over all $\varphi\in C_c^\infty(\mathbb{R}^n)$ with $\|\varphi\|_{L^2(\mathbb{R}^n)}\le 1$ gives
\begin{align*}
\|D_x^\gamma(\chi R_h\psi u)\|_{L^2(\mathbb{R}^n)}\le C_{\gamma,b,M,\chi,\psi}h^M\|u\|_{H^{-b}(\mathbb{R}^n)}.
\end{align*}
Finally, summing over $|\gamma|\le a$ proves
\begin{align*}
\|\chi R_h\psi u\|_{H^a(\mathbb{R}^n)}\le C_{a,b,M,\chi,\psi}h^M\|u\|_{H^{-b}(\mathbb{R}^n)}.
\end{align*}[/guided]