= Solution
Let $X\sim\operatorname{Bin}(n,t)$ and $Y\sim\operatorname{Bin}(n,s)$ be independent. The displayed <Bernstein polynomial> is
$$
B_nf(t,s)=\mathbb E f(X/n,Y/n).
$$
Given $\varepsilon>0$, <uniform continuity> of $f$ supplies $\delta>0$ such that changing both coordinates by less than $\delta$ changes $f$ by less than $\varepsilon$. If $M=\lVert f\rVert_\infty$, then <Chebyshev inequality> and
$\operatorname{Var}(X/n),\operatorname{Var}(Y/n)\leq1/(4n)$ give, uniformly in $(t,s)$,
$$
|B_nf(t,s)-f(t,s)|
\leq\varepsilon+2M\,
\mathbb P\!\left(|X/n-t|\geq\delta
\ \hbox{or}\ |Y/n-s|\geq\delta\right)
\leq\varepsilon+\frac{M}{n\delta^2}.
$$
Thus $B_nf\to f$ uniformly.
Each $B_nf$ is a polynomial, whose iterated integrals may be interchanged term by term. Uniform convergence permits passage to the limit in both iterated integrals, proving the asserted continuous-function form of <Fubini's theorem>.
Now suppose every monomial moment of $f$ vanishes. By linearity,
$\int\!\!\int p f=0$ for every two-variable polynomial $p$. The just-proved Bernstein approximation gives polynomials $p_n\to f$ uniformly. Therefore
$$
\int_{[0,1]^2}f^2
=\lim_n\int_{[0,1]^2}f p_n=0,
$$
so $\boxed{f=0}$. Statement (i) is true.
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