Here's a derivation. It's less clean than I'd like, but it does the job.
Let a triple $(a,b,c)$ denote three visible faces of the die. We'll work modulo 7 and treat the die faces as $\{-3,-2,-1,+1,+2,+3\}$, noting that opposite faces adding to $0$. So $(a,b,c)$ are a permutation of $(\pm 1, \pm2, \pm3)$.
We will express a relationship between three $\pm1$-value functions that describe the triple.
The handedness $h(a,b,c)$ is $+1$ for right-handed triples on the die and $-1$
for left-handed triples.
The sign $s(a,b,c)$ is the number of minus signs among $(a,b,c)$ from
$\{\pm 1, \pm2, \pm3\}$.
The chirality $f(a,b,c)$ is the cyclic order of $(1,2,3)$ in
$(a,b,c)$, ignoring sign. The cyclic order $(1,2,3)$ gives $+1$ and $(1,3,2)$ gives $-1$.
Claim: For any triple, the product of the three function satisfies $h s f = +1$.
Proof: This is satisfied by the triple $(1,2,3)$ for which they are all $+1$. When we negate one of the triple elements, we've mirrored the triple geometrically, so $h$ and $s$ negate but $f$ is sign-independent and stays the same, preserving the equality. When we swap two elements, we negate $h$ and $f$ but keep $s$, again preserving equality. Since these operations suffices to reach any triple, all triples satisfy it.
Now, we use this equality to derive the last element $c$ from the other two elements of the right-handed triple $(a,b,c)$. Right-handed means $h=1$, so we can rewrite the claims as $sf=1$ or $s=f$.
Since $1\times 2 \times 3 = -1 \bmod 7$, we have $s(a,b,c) = -abc$, as each minus sign flips the product.
We can express the chirality $f$ from only $a,b$, since $c$ is forced up to sign, and moreover from $a^2,b^2$, which erases the sign. Squaring mod 7 takes $(1,2,3)$ to $(1,4,2)$, where each element is $4$ times the cyclically previous one modulo 7. So, $b^2/a^2\bmod 7$ is $4$ for this chirality and $2=4^{-1}\bmod 7$ for the opposite one. Therefore, we can find $f$ from the different of this ratio and its inverse:
$$b^2/a^2-a^2/b^2 = 2f \bmod 7$$
and dividing both sides by 2, which is the same as multiplying by -3,
$$-3(b^2/a^2-a^2/b^2) = f \bmod 7$$
Now, plugging expressions into $s=f$ gives
$$abc = 3(b^2/a^2-a^2/b^2). \bmod 7$$
Finally, dividing to isolate $c$ and using the fact the nonzero elements have order 6 modulo 7, so $x^{-3} = x^3$, we get the polynomial expression
$$c=3(a^3b-ab^3)\bmod 7.$$
