It is said in Trace Ideals and Their Applications by Barry Simon that the integral kernel of the operator $g(-i\nabla)$ on $L^2(\mathbb R^n)$ is given by $\breve{g}(x-y)/(2\pi)^n$.
Indeed, for any $\psi\in L^2(\mathbb R^n)$, $$\frac{1}{(2\pi)^n}\int_{\mathbb R^n_y}\breve{g}(x-y)\psi(y)dy=\frac{1}{(2\pi)^n}\int_{\mathbb R^n_y}\int_{\mathbb R^n_k}g(k)e^{ik\cdot (x-y)}\psi(y)dkdy=\frac{1}{(2\pi)^n}\int_{\mathbb R^n_y}\int_{\mathbb R^n_k}g(i\nabla)_ye^{ik\cdot (x-y)}\psi(y)dkdy.$$
Then integration by parts in $y$ gives $$\frac{1}{(2\pi)^n}\int_{\mathbb R^n_y}\int_{\mathbb R^n_k}g(i\nabla)_ye^{ik\cdot (x-y)}\psi(y)dkdy$$
$$=\frac{1}{(2\pi)^n}\int_{\mathbb R^n_y}\Big(\int_{\mathbb R^n_k}e^{ik\cdot (x-y)}dk\Big)g(-i\nabla)\psi(y)dy.$$
Now this is the bit I am confused about. Why are we allowed to use $$\int_{\mathbb R^n_k}e^{ik\cdot (x-y)}dk=(2\pi)^n\delta(x=y).$$ If we can use this, then $$\frac{1}{(2\pi)^n}\int_{\mathbb R^n_y}\Big(\int_{\mathbb R^n_k}e^{ik\cdot (x-y)}dk\Big)g(-i\nabla)\psi(y)dy=\int_{\mathbb R^n_y}\delta(y=x)g(-i\nabla)\psi(y)dy=g(-i\nabla)\psi(x)$$ which shows that $\breve{g}(x-y)/(2\pi)^n$ is the integral kernel of $g(-i\nabla)$.
This is true in the sense of distributions, but it would be great if someone could clarify exactly what sense this is true on $L^2(\mathbb R^n)$.
I will use a slightly different convention of Fourier transform than the one given in Simon's book. It offers you an opportunity to work out the details again as an exercise with Simon's convention.
Here we use the following definition of Fourier transform \begin{align} \hat f(\xi) = \int_{\mathbb{R}^d}e^{-2\pi i\, \xi\cdot x} f(x)\, dx \end{align} and write its inverse transform by $f^{\vee}$. In particular, we have that if $D = \frac{1}{2\pi i}\nabla$, then $\widehat{D f} = \xi\, \hat f$.
By definition, $g(D)$ is defined on $L^2(\mathbb{R}^d)$ as follows \begin{align} (g(D) f)(x):= \left(g(\xi) \hat f(\xi)\right)^\vee = \int_{\mathbb{R}^d}e^{2\pi i\, \xi\cdot x} g(\xi) \hat f(\xi)\, d\xi. \end{align} (If you look carefully at what you have written, you haven't really defined what $g(D)$ actually means.)
Notice that this definition only make sense on a dense subset of $L^2(\mathbb{R}^d)$ since Fourier transform using integral is only defined for $L^1$ function. Anyhow, so if you restrict yourselves to $f$ being a Schwartz function, then we see that \begin{align} \int_{\mathbb{R}^d}e^{2\pi i\, \xi\cdot x} g(\xi) \hat f(\xi)\, d\xi =&\, \int_{\mathbb{R}^d}e^{2\pi i\, \xi\cdot x} g(\xi) \left(\int_{\mathbb{R}^d} e^{-2\pi i\, y\cdot \xi}f(y)\, dy\right)d\xi\\ =&\, \int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d} e^{-2\pi i\, (y-x)\cdot \xi}g(\xi)\ d\xi\right)f(y)\, d y\\ =&\, \int_{\mathbb{R}^d} g^\vee (x-y)f(y)\, d y \end{align} which is the desired result.
Additional: To answer the other question in the comment, let us consider the operator $A = h(x)g(-i\nabla)$.
First, we need to understand how $A$ is defined. Well, it should be clear that \begin{align} (A f)(x) := h(x) (g(-i\nabla) f)(x) \end{align} i.e. apply the operator $g(-i\nabla)$ then apply the multiplication operator $h(x)$.
As above, we see that \begin{align} (g(-i\nabla) f)(x)= \int_{\mathbb{R}^d} g^\vee(x-y) f(y)\, dy \end{align} which means \begin{align} (A f)(x) =&\, h(x)\int_{\mathbb{R}^d} g^\vee(x-y) f(y)\, dy\\ =&\, \int_{\mathbb{R}^d} \underbrace{h(x)g^\vee(x-y)}_{k(x, y)} f(y)\, dy. \end{align} As pointed out by you, that the kernel should be \begin{align} k(x, y) = h(x)g^\vee(x-y). \end{align}
Now, let us address your question, "what sense would $h(x)g^\vee(x-y)$ be the integral kernel for $A$?"
First, the theorem that you are quoting is telling you when $A$ a Hilbert--Schmidt operator. Of course, you can not expect every $h(x)g(-i\nabla)$ to be Hilbert--Schmidt. For instance $h(x)g(-i\nabla) = -x\, \Delta$ which is clearly unbounded therefore can't possibly be a compact operator (Hilbert--Schmidt are compact operators). Nevertheless, you could write out its integral kernel \begin{align} -x\, \Delta f =&\, -x \int_{\mathbb{R}^d} \Delta_x \delta(x-y) f(y)\, dy \\ =&\, -x \int_{\mathbb{R}^d} \Delta_x \left(\int_{\mathbb{R}^d} e^{2\pi i (x-y)\cdot\xi}\, d\xi\right) f(y)\, d y\\ =&\, -x \int_{\mathbb{R}^d} \underbrace{\left( \int_{\mathbb{R}^d} e^{2\pi i (x-y)\cdot\xi}\underbrace{(-4\pi^2|\xi|^2)}_{g(\xi)}\, d\xi\right)}_{g^\vee(x-y)} f(y)\, d y. \end{align}
The point of this example is to show you that $g$ doesn't even have to be integrable. Of course, what this means is that the Fourier (or inverse Fourier) transform of $g$ is defined rigorously in the sense of Fourier transform of tempered distributions.
Back to the theorem: $A$ is a Hilbert--Schmidt operator if and only if $h, g$ are $L^2$ integrable functions.
Hence $A=-x\Delta$ is not a Hilbert--Schmidt operator but \begin{align} A:= \frac{1}{(1+|x|)^{d+2}}\frac{1}{(1+|-i\nabla|)^{d+2}} \end{align} is a Hilbert--Schmidt operator.