Here $A$, $B$, $C$ are arbitrary $n\times n $ matrices. I can find a solution as
$$A^2 + B^2 + C^2 = ((A+ iB )(A-iB) + (A-i B) (A+iB))/2 + C^2 ,$$
and identify $E = (A+ iB)/\sqrt{2}$, $F =( A- iB)/\sqrt{2}$, and $G=H = C/\sqrt{2}$.
Is it possible to avoid the imaginary unit $i$? That is, is there a solution such that $E,F,G,H$ are real combinations of $A,B,C$?
ps. The original problem is actually the following. I have a $3\times 3 $ complex symmetric matrix $M$. By the Autonne–Takagi factorization, it can be written as
$$M = u u^T + v v^T + w w^T ,$$
where $u,v,w$ are three column vectors. I want to rewrite $M $ as
$$ M = (e f^T+ f e^T ) + (g h^T + h g^T ), $$
with four column vectors $e, f, g, h$. Or in other words, I have a symmetric function $M(x_1, x_2)$, with $ x_i = 1, 2, 3$. I want to find functions $e(x), f(x), g(x),h(x)$ such that
$$M (x_1, x_2) = f(x_1)g(x_2) + g(x_1) f(x_2) + g(x_1)h(x_2) + h(x_1) g(x_2). $$
The idea is actually that, taking two points on the manifold of the functions $\{ f(x_1)g(x_2) + g(x_1) f(x_2) \}$, we can span the whole space of symmetric matrices.
I'll address the problem stated at the beginning of your question rather than your original problem.
Suppose there exists an algebraic identity \begin{align} A^2+B^2+C^2 &=(x_1A+y_1B+z_1C)(x_2A+y_2B+z_2C)\\ &+(x_2A+y_2B+z_2C)(x_1A+y_1B+z_1C)\\ &+(x_3A+y_3B+z_3C)(x_4A+y_4B+z_4C)\\ &+(x_4A+y_4B+z_4C)(x_3A+y_3B+z_3C)\tag{1} \end{align} where the $x_i$s, $y_i$s and $z_i$s are constant real numbers that does not depend on $A,B$ and $C$. When the matrices are $3\times3$ or larger, since there exist $A,B,C$ such that $$ AB\ne0=A^2=B^2=C^2=BA=AC=CA=BC=CB, $$ the coefficients of $AB$ on both sides of $(1)$ must be equal. Similarly, as there exist $A,B,C$ such that $$ A^2\ne0=B^2=C^2=AB=BA=AC=CA=BC=CB, $$ the coefficients of $AB$ on both sides of $(1)$ must be equal. It follows that, by permuting the roles of $A,B$ and $C$, we can compare coefficients of each of $A^2,B^2,C^2,AB,BA,AC,CA,BC$ or $CB$ on both sides of $(1)$. The problem thus reduces to solving the system of equations \begin{cases} 2(x_1x_2+x_3x_4)=1,\\ 2(y_1y_2+y_3y_4)=1,\\ 2(z_1z_2+z_3z_4)=1,\\ x_1y_2+x_2y_1+x_3y_4+x_4y_3=0,\\ x_1z_2+x_2z_1+x_3z_4+x_4z_3=0,\\ y_1z_2+y_2z_1+y_3z_4+y_4z_3=0. \end{cases} (Since $AB$ and $BA$ have the same coefficients and similarly for other pairs of cross terms, there are only six equations in total.) You may try to solve this using a computer algebra system.