(Definitions & set up:) Let $H_1,\dots,H_N$ be Hilbert spaces with orthonormal bases $B_1,\dots,B_n, e_k^{j}\in B_k, j \in \mathbb{Z}$ and $H = \prod_{k=1}^NH_k$. Write $I_k$ as the index set of the ONB $B_k$ and define $I = \prod_{k=1}^NI_k$.
(*) Suppose that $T, U:H\to \mathbb{C}$ are separately conjugate linear and continuous in each argument slot $k = 1,\dots,N$ and that $\sum_{l\in I}|T(e_l)|^2 \langle \infty$ with $e_l = (e_1^{l_1},\dots,e_N^{l_N})\in \prod_{k=1}^N B_k \equiv B$, and the same for $U$. Then by lecture notes define the operaration $\langle .,.\rangle $ (which turns out to be an inner product) between $T$ and $U$ as
(**) $$\langle T,U\rangle \equiv \sum_{l\in I}T(e_l)^*U(e_l)$$
where $^*$ is the complex conjugate.
Suppose that $T:L^2([0,1]^n)\otimes L^2([0,1]^m)\to L^2([0,1]^{n + m})$ is an unitary map, i.e. an isomorphism of the two spaces. One such unitary map is given by $T(e_1^{l_1}\otimes e_2^{l_2})(x_1,x_2) = e_1^{l_1}(x_1)e_2^{l_2}(x_2), l \in \mathbb{Z}^{n+m}, l_1\in \mathbb{Z}^n, l_2\in\mathbb{Z}^m$
Note that elements of $L^2([0,1]^n)\otimes L^2([0,1]^m)$ and $L^2([0,1]^{n + m})$ are functions which satisfy the conditions of (*) for their respective spaces. Thus when computing $\langle f, g\rangle $ of two such elements of the same space we should in principle be able to use the formula of (**).
During one presented proof of my notes, the author claims that
$$\langle T(e_1^{l_1}\otimes e_2^{l_2}), T(e_1^{\hat{l}_2}\otimes e_2^{\hat{l}_2})\rangle = \int_{[0,1]^{n+m}}(e_1^{l_1}(x_1)e_2^{l_2}(x_2))^*(e_1^{\hat{l}_1}(x_1)e_2^{\hat{l}_1}(x_2))dx$$
This makes sense to me in the sense that by definition of the mapping $T$
$$\langle T(e_1^{l_1}\otimes e_2^{l_2}), T(e_1^{\hat{l}_2}\otimes e_2^{\hat{l}_2})\rangle = \langle e_1^{l_1}e_2^{l_2}, e_1^{\hat{l}_1}e_2^{\hat{l}_2}\rangle $$
and I can believe the identity as per how inner products usually work in $L^p$ spaces.
(Question:) But what is not immediately clear to me is how one could jump from the continuous integral to the sum formula of (**) since $T(e_1^{l_1}\otimes e_2^{l_2})$ and $T(e_1^{\hat{l}_2}\otimes e_2^{\hat{l}_2})$ are functions satisfying (*). Moreover, to my knowledge $L^2([0,1]^{n+m})$ is separable so that the ONB is countable, while taking an integrable over $[0,1]^{n+m}$ surely is an uncountable sum.
It could be that I am missing something and the two representations do agree. Thanks!
The fact that $T$ is unitary immediately implies that $$ \langle T(e_1^{l_1}\otimes e_2^{l_2}), T(e_1^{\hat{l}_2}\otimes e_2^{\hat{l}_2})\rangle = \langle e_1^{l_1}\otimes e_2^{l_2}, e_1^{\hat{l}_2}\otimes e_2^{\hat{l}_2}\rangle. $$ And, as you say, by definition of $T$ and of the inner product on $L^2(X,\mu)$, i.e. $\langle f,g\rangle=\int_X f\overline g\,d\mu$, \begin{align} \langle T(e_1^{l_1}\otimes e_2^{l_2}), T(e_1^{\hat{l}_2}\otimes e_2^{\hat{l}_2})\rangle &= \langle e_1^{l_1}e_2^{l_2}, e_1^{\hat{l}_1}e_2^{\hat{l}_1}) \rangle \\[0.3cm] &= \int_{[0,1]^{n+m}}(e_1^{l_1}(x_1)e_2^{l_2}(x_2))^*(e_1^{\hat{l}_1}(x_1)e_2^{\hat{l}_1}(x_2))dx. \end{align} The above is completely standard, and is maybe obscured by the notation. But the $T$ you defined and I used here, does not satisfy $(*)$. The $T$ in $(*)$ sends elements of $H$ to scalars.