I don't think there may be deep connections hereThere is no way to translate mere orthogonality into independence or even into a martingale condition. The
Indeed, the independence of real-valued random variables (r.v.'s) $X_1,\dots,X_N$ is a very strong condition, involving a continuum equations, say $P(X_1\le x_1,\dots,X_N\le x_N)=P(X_1\le x_1)\cdots P(X_N\le x_N)$ for all $(x_1,\dots,x_N)\in\mathbb R^N$.
The condition that $X_1,\dots,X_N$ are martingale differences is less restrictive than the independence (given that the $X_i$'s are zero-mean), but it still involves a continuum equations, including (say) $E(X_i|X_{i-1}=x_{i-1})=0$ for all $i=2,\dots,N$ and all $(x_1,\dots,x_{N-1})\in\mathbb R^{N-1}$.
On the other hand, the orthogonality of $X_1,\dots,X_N$ is a much, much weaker condition, involving only finitely many (namely, $\binom N2$) equations.
On the other hand, the r.v.'s $X_k:=e^{ikU}$, where $k=1,2,\dots$ and $U$ is a r.v. uniformly distributed on $[0,2\pi)$, which are implicitly used in Fourier analysis, are not merely mutually orthogonal. In particular, here we have the much stronger condition $E\prod_{j=1}^N X_{k_j}^{p_j}=0$ whenever $\sum_{j=1}^N {p_j}{k_j}\ne0$, which is somewhat close to the stronger condition $E\prod_{j=1}^N Y_{k_j}^{p_j}=0$ whenever $\sum_{j=1}^N ({p_j}{k_j})^2\ne0$, where the $p_k$'s and $k_j$'s are integers, $Y_k:=e^{ikU_k}$, and the $U_k$'s are independent copies of the r.v. $U$.