= Solution
Expansion of the two squares gives
$$
\boxed{|\xi^+|^2+|\xi^-|^2=\frac12(|\xi|^2+|\xi|^2|\sigma|^2)=|\xi|^2}.
$$
Let $D=d(f,g)$ be the <Fourier distance of order two>, with the supremum taken over $\xi\ne0$. Both <Fourier transforms> have modulus at most one. Add and subtract $\widehat g(\xi^+)\widehat f(\xi^-)$, then use the <triangle inequality>:
$$
|\widehat f(\xi^+)\widehat f(\xi^-)-\widehat g(\xi^+)\widehat g(\xi^-)|
\leq|\widehat f(\xi^+)-\widehat g(\xi^+)|+|\widehat f(\xi^-)-\widehat g(\xi^-)|
\leq D(|\xi^+|^2+|\xi^-|^2).
$$
Dividing by $|\xi|^2$ proves the bound by $D$, and splitting the last expression into its two weighted terms gives the requested intermediate inequality. If one of $\xi^\pm$ vanishes, its unweighted difference is zero by equal mass; its weighted term is interpreted as zero, avoiding a $0/0$ quotient.
Equal mass and first moment also explain finiteness of this distance: subtract the constant and linear Taylor terms in the Fourier <integral> and use $|e^{-ia}-1+ia|\leq a^2/2$. This bounds the difference by $|\xi|^2\int|v|^2(f+g)/2$. No direction-independent extension of the quotient at zero is required.
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