A simpler formula for the map of $V_{2n}\times V_{2n}\to V_{4n-2}$ is this: You just send $F\wedge G$ to $\mathrm{d}F\wedge\mathrm{d}G$ which is a $2$-form in the variables $x_1,x_2$ and then take the coefficient of the basis $2$-form $\mathrm{d}x_1\wedge\mathrm{d}x_2$. In other words, you send $F\wedge G$ to the Poisson bracket $\{F,G\}$.

Yes. That's correct.

Yes. Using the $\mathrm{G}_2$-invariant almost complex structure, there is a $\mathrm{G}_2$-invariant $(0,3)$-form, and that corresponds to the killing spinor under the above isomorphisms. The reason is that $S^6=\mathrm{G}_2/\mathrm{SU}(3)$, and $\mathrm{SU}(3)$ is the stabilizer of a $(0,3)$-form (unique up to a complex constant multiple).

@emiliocba: I think you are correct about the sign of $\phi_3$. I didn't actually check it when I wrote the answer back then. The OP's 'standard basis' is different from mine, that's why I suggested that the OP check. I like my basis better. (I would use $e_7 = i(jl)$ instead of $e_7=(ij)l$ because then left multiplication by $e_1=i$ induces the 'correct' complex structure on $e_1^\perp\simeq \mathbb{R}^6 = \mathbb{C}^3$.)

@PaulCusson: An Hermitian connection $A$ on the tangent bundle that is without torsion wil always have $F_A^{0,2}=0$. This is a consequence of the first Bianchi identity.

@MatheusAndrade: Hmmm. You are right that something is wrong with my answer, but I'm not sure what. I'm traveling now and don't have those calculations with me so I can't check them. I'll have a look at them when I get back home next week, check them over, and see what I can do about correcting the formulae.

Moreover, in dimension $3$, if $f(x)$ is nonvanishing, then $R_{ijkl}=f(x)(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})$ is locally the Riemann curvature tensor of some metric $g$. I proved this when $f$ is real-analytic (and nonvanishing) and later DeTurck and Yang proved this when $f$ is smooth (and nonvanishing).

Actually, this doesn't quite work. There are non constant functions $h(x)$ such that $g = h(x)\,\bigl (\mathrm{d}x^1)^2 + \cdots + (\mathrm{d}x^n)^2\bigr)$ has constant sectional curvature. For example, $h(x) = 4/(1+|x|^2)^2$. Won't the Riemann curvature tensor of $g$ have the above form for a nonconstant $f(x)$?

Oh, I wasn't worried about 'attribution'. I was just pointing out that it's not as 'concrete' as the OP might want. What I was thinking was that the OP might just want a 'concrete' answer, such as $R = x^3\,(\mathrm{d}x^1\wedge\mathrm{d}x^2)\otimes (\mathrm{d}x^1\wedge\mathrm{d}x^2)$ on $\mathbb{R}^3$.

Your example is basically what I gave in my answer that the OP linked. It may be that the OP will not consider your class of examples sufficiently 'explicit' either.

@DanielCastro: Well, an area bound won' t work because you could just take $\Omega_1$ to be the unit disk and $\Omega_2$ to be an ellipse with a large major axis and sufficiently small minor axis, and you'd have the area bound you propose, but the same argument as above for circles would show that the desired $f$ won't exist. Maybe if you assume that $\Omega_1$ and $\Omega_2$ are convex and the '$XY$-diameter' of $\Omega_2$ is sufficiently small relative to the '$xy$-diameter' of $\Omega_1$, you'd have a chance, but I suspect that it's still hopeless with only these hypotheses.