Yes, if the quartic $C : F_4=0$ is smooth then the line $l : f_1=0$ must be one of the $28$ bitangents (because the restriction $F_4|_l$ is the square of $f_2|_l$), and then $f_2$ restricted to $l$ is one of the square roots of $F_4|_l$, and each of the lifts of $f_2|_l$ lets you solve for $f_3$.
Note that writing $F_4$ as $f_1 f_3 - f_2^2$ is not a factorization, and a parameter count would already indicate that there should be infinitely many such expressions: there are $3+6+10 = 19$ undetermined coefficients in $f_1,f_2,f_3$, and only $15$ coefficients to match in $F_4$. But there's also a $4$-dimensional group acting: multiply $f_1,f_3$ by $\lambda,\lambda^{-1}$, or change $f_2,f_3$ to $f_2 + g f_1$ and $f_3 + 2 g f_2 + g^2 f_1$ for any linear form $g$. So we expect finitely many solutions up to these transformations, and indeed we get the $28$ pairs described in the first paragraph.
(Likewise for $F_4 = b^2-ac$ with $a,b,c$ each of degree $2$ we have $2^6 - 1 = 63$ three-dimensional families, indexed by the nontrivial $2$-torsion divisor classes $D$ in the Jacobian of $C$; the function $a/b=b/c$ on $F_4$ is the quotient of two sections of $D+K$ where $K$ is the canonical divisor.)