I think this is not possible.

Call $f(x,y)$ the characteristic function of the set $\{x<y\}$ in $R^2$. If $h(y)$ is any function of the single variable $y$, we have $\sup_y|f(x,y)-h(y)|\ge 1/2$ for a.e. $x$. Now, suppose you can approximate $f$ (locally) with tensor products; then you can approximate $f$ with tensor products of simple functions. Let $u(x,y)=\sum g_i(x)h_i(x)$ be any such function. Represent all $g_i$ in the form $g_i(x)=\sum_{k=1}^N c_{ik}\chi_{E_k}$ with the same $N$ and the same sets $E_k$. You see that for $x\in E_1$ we have $u(x,y)=\sum c_{1k}h_k(y)$ independet of $x$, hence $\sup_y|f(x,y)-u(x,y)|\ge 1/2$ on $E_1$. The same argument applies to all $E_i$ and we have a contradiction.

I think this is not possible.

Call $f(x,y)$ the characteristic function of the set $\{x<y\}$ in $R^2$. If $h(y)$ is any function of the single variable $y$, we have $\sup_y|f(x,y)-h(y)|\ge 1/2$ for a.e. $x$. Now, suppose you can approximate $f$ (locally) with tensor products; then you can approximate $f$ with tensor products of simple functions. Let $u(x,y)=\sum g_i(x)h_i(y)$ be any such function. Represent all $g_i$ in the form $g_i(x)=\sum_{k=1}^N c_{ik}\chi_{E_k}$ with the same $N$ and the same sets $E_k$. You see that for $x\in E_1$ we have $u(x,y)=\sum c_{1k}h_k(y)$ independent of $x$, hence $\sup_y|f(x,y)-u(x,y)|\ge 1/2$ on $E_1$. The same argument applies to all $E_i$ and we have a contradiction.