Unfortunately, taking $\mathcal{H}=H^2(\mathbb{R}^3)$ won't work. If $f\in H^2(\mathbb{R}^3)$ then, in general, it is not true that $\Delta f\in H^2(\mathbb{R}^3)$. This is a problem, because the Hamiltonian must be an operator from $\mathcal{H}$ to $\mathcal{H}$.
The only mathematically sound way out is taking $\mathcal{H}=L^2(\mathbb{R}^3)$ and considering the Laplacian as an unbounded operator, meaning that it is not defined on the whole $L^2(\mathbb{R}^3)$ space but only on a dense subspace. This idea dates back to von Neumann.
There's actually no problem, because you're looking at functions of the form $f(x_1,x_2) = \Psi_1(x_1)\Psi_2(x_2)$, so that
$$
\mathcal{F}\{\Psi_1(x_1)\Psi_2(x_2)\}(k_1,k_2) = \int_{\mathbb{R}^{n_1+n_2}}\Psi_1(x_1)\Psi_2(x_2)e^{2\pi i (x_1,x_2) \cdot (k_1,k_2)} d^{n_1}x_1 d^{n_2}x_2\\
= \left(\int_{\mathbb{R}^{n_1}} \Psi_1(x_1) e^{2\pi i x_1 \cdot k_1} d^{n_1}x_1\right)\left(\int_{\mathbb{R}^{n_2}} \Psi_2(x_2) e^{2\pi i x_2 \cdot k_2} d^{n_2}x_2\right)\\
= \mathcal{F}\{\Psi_1(x_1)\}(k_1) \mathcal{F}\{\Psi_2(x_2)\}(k_2),
$$
as you want.
Abstractly, to clarify L&L's abuse of notation, what's really going on is that the composite of non-interacting Hamiltonians
- $\hat{H}_1$ on $L^2(\mathbb{R}^{n_1},d^{n_1}x_1)$,
- $\hat{H}_2$ on $L^2(\mathbb{R}^{n_2},d^{n_2}x_2)$,
is given by $\hat{H}_1 \otimes I + I \otimes \hat{H}_2$ on
$$
L^2(\mathbb{R}^{n_1},d^{n_1}x_1) \otimes L^2(\mathbb{R}^{n_2},d^{n_2}x_2) \cong L^2(\mathbb{R}^{n_1+n_2},d^{n_1}x_1 d^{n_2}x_2),
$$
and that the Fourier transform
$$
\mathcal{F} : L^2(\mathbb{R}^{n_1+n_2},d^{n_1}x_1 d^{n_2}x_2) \cong L^2(\mathbb{R}^{n_1},d^{n_1}x_1) \otimes L^2(\mathbb{R}^{n_2},d^{n_2}x_2)\\ \to L^2(\mathbb{R}^{n_1},(2\pi)^{-n_1/2}d^{n_1}k_1) \otimes L^2(\mathbb{R}^{n_2},(2\pi)^{-n_2/2}d^{n_2}k_2) \cong L^2(\mathbb{R}^{n_1+n_2},(2\pi)^{-(n_1+n_2)/2}d^{n_1}k_1 d^{n_2}k_2)
$$
is given by $\mathcal{F}_1 \otimes \mathcal{F}_2$, where $\mathcal{F}_j$ is the Fourier transform
$$
\mathcal{F}_j : L^2(\mathbb{R}^{n_j},d^{n_j}x_j) \to L^2(\mathbb{R}^{n_j},(2\pi)^{-n_j/2}d^{n_j}k_j)
$$
so that
$$
\mathcal{F} (\hat{H}_1 \otimes I + I \otimes \hat{H}_2) \mathcal{F}^{-1} = (\mathcal{F}_1 \hat{H}_1 \mathcal{F}_1^{-1}) \otimes I + I \otimes (\mathcal{F}_2 \hat{H}_2 \mathcal{F}_2^{-1})
$$
on
$$
L^2(\mathbb{R}^{n_1},(2\pi)^{-n_1/2}d^{n_1}k_1) \otimes L^2(\mathbb{R}^{n_2},(2\pi)^{-n_2/2}d^{n_2}k_2) \cong L^2(\mathbb{R}^{n_1+n_2},(2\pi)^{-(n_1+n_2)/2}d^{n_1}k_1 d^{n_2}k_2).
$$
Given all this, what L&L are saying (modulo abuse of notation) is that if
$$
\hat{H}_1 f_1(x_1) = E_1 f_1(x_1), \quad \hat{H}_2 f_2(x_2) = E_2 f_2(x_2),
$$
then $(f_1 \otimes f_2)(x_1,x_2) := f_1(x_1)f_2(x_2)$ satisfies
$$
(\hat{H}_1 \otimes I + I \otimes \hat{H}_2)(f_1 \otimes f_2)(x_1,x_2) = (E_1+E_2)(f_1 \otimes f_2)(x_1,x_2);
$$
using $\mathcal{F} = \mathcal{F}_1 \otimes \mathcal{F}_2$, then, you get that
$$
\mathcal{F}(f_1 \otimes f_2)(k_1,k_2) = \mathcal{F}_1\{f_1\} \otimes \mathcal{F}_2\{f_2\}(k_1,k_2) = \mathcal{F}_1\{f_1\}(k_1)\mathcal{F}_2\{f_2\}(k_2),
$$
as you wanted.
Best Answer
The answer to my question is Duhamel's principle.