You may know this already, but, if not, the following may be helpful to you:

The ring of binary invariants for sextic polynomials is generated in degrees 2, 4, 6, 10, and 15. (See this reference on binary sextics.) Thus, in your notation (at least when the field $\mathbb{F}$ has characteristic $0$) for $k=2,4,6,10, 15$ there exists a homogeneous of degree $k$ polynomial mapping $Q_k:S^6(\mathbb{F}^2)\to \mathbb{F}$ that satisfies $$ Q_k\bigl(F_U(x,y)\bigr) = \det(U)^{3k}\,Q_k\bigl(F(x,y)\bigr) $$ for all $U\in\mathrm{GL}(2,\mathbb{F})$, and, moreover, every polynomial $Q: S^6(\mathbb{F}^2)\to \mathbb{F}$ that satisfies $$ Q\bigl(F_U(x,y)\bigr) = Q\bigl(F(x,y)\bigr) $$ for all $U\in\mathrm{SL}(2,\mathbb{F})$ is, in fact, a polynomial in $1,Q_2,Q_4,Q_6,Q_{10}, Q_{15}$. (Note that ${Q_{15}}^2$ is actually a polynomial in $Q_2,Q_4,Q_6,Q_{10}$, and this is the only relation between these polynomials.)

Taking $U = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$, we see that when $F(x,y)$ can be written in the desired form, we have $F_U(x,y) = F(x,y)$, so we must have $$ Q_{15}\bigl(F(x,y)\bigr) = 0. $$ The polynomial $Q_{15}$ is, of course, irreducible, since, if it were to factor, one of the factors would have to be an invariant of lower odd degree, and this is known not to exist. Thus, the hypersurface $Q_{15}=0$ in $S^6(\mathbb{F})$ must be an irreducible component of the hypersurface $L\subset S^6(\mathbb{F})$ consisting of sextics that can be written in the desired form after a linear change of variables.

On the other hand, it seems that $L$ is likely to be irreducible, since, over $\mathbb{C}$ anyway, we know, by Noam's observation/comment that it can be 'parametrized' by a connected algebraic variety. Assuming this detail, it would follow that the necessary and sufficient condition for $F(x,y)$ to be written in the above form using complex coefficients, is that it lie in the zero locus of $Q_{15}$. (I don't know how to tell when it can be done with real coefficients.)

You may know this already, but, if not, the following may be helpful to you:

The ring of invariants for binary sextics is generated in degrees 2, 4, 6, 10, and 15. (See this reference on binary sextics.) Thus, in your notation (at least when the field $\mathbb{F}$ has characteristic $0$) for $k=2,4,6,10, 15$ there exists a polynomial $Q_k:S^6(\mathbb{F}^2)\to \mathbb{F}$ that is homogeneous of degree $k$ that satisfies $$ Q_k\bigl(F_U(x,y)\bigr) = \det(U)^{3k}\,Q_k\bigl(F(x,y)\bigr)\tag1 $$ for all $U\in\mathrm{GL}(2,\mathbb{F})$, and, moreover, every polynomial $Q: S^6(\mathbb{F}^2)\to \mathbb{F}$ that satisfies $$ Q\bigl(F_U(x,y)\bigr) = Q\bigl(F(x,y)\bigr) $$ for all $U\in\mathrm{SL}(2,\mathbb{F})$ is, in fact, a polynomial in $\{1,Q_2,Q_4,Q_6,Q_{10},Q_{15}\}$. It is known that ${Q_{15}}^2$ is a polynomial in $\{Q_2,Q_4,Q_6,Q_{10}\}$ and that this is the only relation among these polynomials. Consequently, the polynomials $Q_{k}$ are irreducible.

Taking $U = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$, one sees that when $F(x,y)$ is in the desired form, one has $F_U(x,y) = F(x,y)$. Conversely, $F(x,y)$ can be put into the desired form over the algebraic closure $\mathbb{F}^+$ of $\mathbb{F}$ if and only if there exists a $U\in\mathrm{GL}(2,\mathbb{F}^+)$ satisfying $U^2 = I$ and $U\not=I$ such that $F_U(x,y) = F(x,y)$. Using (1) with $k=15$, one finds that the existence of such a $U$ for a given $F(x,y)$ implies $$ Q_{15}\bigl(F(x,y)\bigr) = 0. $$ Thus, the hypersurface $Q_{15}=0$ in $S^6(\mathbb{F})$ must be an irreducible component of the hypersurface $L\subset S^6(\mathbb{F})$ consisting of sextics that can be written in the desired form after a linear change of variables.

Meanwhile, it seems likely that $L$ is irreducible, since, over $\mathbb{F}^+$ anyway, we know, by Noam's observation/comment that it can be 'parametrized' by a connected algebraic variety. Assuming this detail, it would follow that the necessary and sufficient condition for $F(x,y)$ to be written in the above form over $\mathbb{F}^+$ is that it lie in the zero locus of $Q_{15}$. (I don't know how to tell when it can be done over $\mathbb{F}$, but perhaps this can be deduced by knowing something about the values $Q_k\bigl(F(x,y)\bigr)$ for $k=2,4,6,10$.)