Question on finding a Hilbert space isomorphism between $L^2([0,R]^n)\otimes L^2([0,R]^m)$ and $L^2([0,R]^{n+m})$

Question

Let $R > 0, n,m\in\mathbb{N}$ and suppose that we have already showed that the map $f:L^2([0,R]^n)\otimes L^2([0,R]^m)\to L^2([0,R]^{n+m})$ is a linear bijective isometry between the basis vectors of $L^2([0,R]^n)\otimes L^2([0,R]^m)$ and $L^2([0,R]^{n+m})$ with $f(e_i\otimes e_j)(x,y) := e_i(x)e_j(y)$. How could I then extend $f$ to act on any element of $L^2([0,R]^n)\otimes L^2([0,R]^m)$? We have an explicit representation of $f(\alpha)$ when $\alpha$ is any finite linear combination of the basis vectors $e_i\otimes e_j$. But how do we take the leap to the case of $\alpha$ being a countably infinite sum of $e_i\otimes e_j$s, i.e. when we represent elements of $L^2([0,R]^n)$ and $L^2([0,R]^m)$ with their Fourier series?