I've always thought of this formula as an amusing accident. You say "I can go through the proofs", so I don't know if what I'll say helps any more than that, but here goes.

1. If $M$ is any endomorphism, it's easy to show $Trace(Sym^2(M)) = (Trace(M)^2 + Trace(M^2))/2$. Proof: check it for diagonal $M$, obtain for diagonalizable, extend by continuity to all. The same technique gives $Trace(Alt^2(M)) = (Trace(M)^2-Trace(M^2))/2$.

2. Basic character theory fact: the average of $Trace(g|_V)$ is $\dim V^G$.

If $V$ is a real representation, and $G$ is compact (yours being finite, apparently), then we can average a bilinear positive definite symmetric inner product to get a $G$-invariant one, which gives an isomorphism $V \equiv V^*$. That passes to the complexification $V_{\mathbb C}$. Hence there is an invariant vector (the inner product) in $ S ym^2(V)$.

If $V$ is a quaternionic representation, and $G$ is compact, then I admit I forget how to get an antisymmetric form on the complex representation $Forget(V)$. Maybe one can average a quaternionic-Hermitian form to get a $G$-invariant one?

Moreover one wants the converses of these: having the symmetric or antisymmetric inner products lets one realize a complex irrep as a complexification or $Forget()$ of a quaternionic rep. I don't remember this being too hard.

Then come a couple of minor miracles. Since $V$ is irreducible, Schur's lemma says there is at most a $1$-d space of invariant maps $V \to V^*$, i.e. $G$-invariant vectors in $V\otimes V$. Since $V\otimes V$ is a $G\times Z_2$-rep, any invariant is either in $Sym^2(V)$ or $Alt^2(V)$, not some weird mix. Hence there are only three cases: no invariant at all, $\dim (Sym^2(V))^G = 1$ and $\dim (Alt^2(V))^G = 0$, or vice versa. We can figure out which occurs by looking at $\dim (Sym^2(V))^G - \dim (Alt^2(V))^G = 0,1,-1$. By facts 1 & 2 above, that's the F-S indicator.

I've always thought of this formula as an amusing accident. You say "I can go through the proofs", so I don't know if what I'll say helps any more than that, but here goes.

1. If $M$ is any endomorphism, it's easy to show $Trace(Sym^2(M)) = (Trace(M)^2 + Trace(M^2))/2$. Proof: check it for diagonal $M$, obtain for diagonalizable, extend by continuity to all. The same technique gives $Trace(Alt^2(M)) = (Trace(M)^2-Trace(M^2))/2$.

2. Basic character theory fact: the average of $Trace(g|_V)$ is $\dim V^G$.

If $V$ is a real representation, and $G$ is compact (yours being finite, apparently), then we can average a bilinear positive definite symmetric inner product to get a $G$-invariant one, which gives an isomorphism $V \equiv V^*$. That passes to the complexification $V_{\mathbb C}$. Hence there is an invariant vector (the inner product) in $ S ym^2(V)$.

If $V$ is a quaternionic representation, and $G$ is compact, then I admit I forget how to get an antisymmetric form on the complex representation $Forget(V)$. Maybe one can average a quaternionic-Hermitian form to get a $G$-invariant one?

Moreover one wants the converses of these: having the symmetric or antisymmetric inner products lets one realize a complex irrep as a complexification or $Forget()$ of a quaternionic rep. I don't remember this being too hard.

Then come a couple of minor miracles. Since $V$ is irreducible, Schur's lemma says there is at most a $1$-d space of invariant maps $V \to V^*$, i.e. $G$-invariant vectors in $V \otimes V$. Since $V\otimes V$ is a $G\times Z_2$-rep, any $G$-invariant is either in $Sym^2(V)$ or $Alt^2(V)$, not some weird mix. Hence there are only three cases: no invariant at all, $\dim (Sym^2(V))^G = 1$ and $\dim (Alt^2(V))^G = 0$, or vice versa. We can figure out which occurs by looking at $\dim (Sym^2(V))^G - \dim (Alt^2(V))^G = 0,1,-1$. By facts 1 & 2 above, that's the F-S indicator.