= Solution
Put
$$
c=y\cdot z,\qquad p=x\cdot y,\qquad q=x\cdot z.
$$
Since $x,y,z$ are unit <vectors>,
$$
S=2-p^2-q^2+p+q
=\frac52-\left(p-\frac12\right)^2
-\left(q-\frac12\right)^2.
$$
The <feasible inner products with two unit vectors> satisfy
$$
\frac{p^2-2cpq+q^2}{1-c^2}\leq1.
$$
The desired pair $p=q=1/2$ is feasible exactly when $c\geq-1/2$. In that case every maximizer has these <inner products>. If $c<-1/2$, symmetry and strict convexity force $p=q$, and the nearest feasible diagonal point is
$$
p=q=\sqrt{\frac{1+c}{2}}.
$$
Using the <vector triple product>,
$$
x\times(y\times z)=y(x\cdot z)-z(x\cdot y)=qy-pz.
$$
Hence
$$
\boxed{
F(y,z)=
\begin{cases}
\dfrac12(y-z),&y\cdot z\geq-\dfrac12,\\[6pt]
\sqrt{\dfrac{1+y\cdot z}{2}}\,(y-z),
&y\cdot z<-\dfrac12.
\end{cases}}
$$
Rotations preserve dot products and cross-product lengths, while $z'=2z$. Thus
$$
S'=5-p^2-4q^2+p+2q
=\frac{11}{2}
-\left(p-\frac12\right)^2
-4\left(q-\frac14\right)^2.
$$
Let
$$
t=\begin{pmatrix}1/2\\1/4\end{pmatrix},
\qquad
K=\begin{pmatrix}1&c\\c&1\end{pmatrix},
\qquad
W=\begin{pmatrix}1&0\\0&4\end{pmatrix}.
$$
The target pair is feasible precisely when
$$
t^TK^{-1}t=\frac{5-4c}{16(1-c^2)}\leq1,
$$
or
$$
\frac{1-3\sqrt5}{8}\leq c\leq
\frac{1+3\sqrt5}{8}.
$$
On this interval,
$$
\boxed{G(y,z)=\frac14y-\frac12z}.
$$
Outside this interval, let $\lambda>0$ be the unique number for which
$$
r=(W+\lambda K^{-1})^{-1}Wt,
\qquad
r^TK^{-1}r=1,
$$
and write $r=(p,q)^T$. These equations give the unique weighted projection of $t$ onto the feasible ellipse, and therefore determine the maximizing <inner products> solely from $c=y\cdot z$. In the remaining cases,
$$
\boxed{G(y,z)=qy-pz}.
$$
Neither answer depends on $\theta_1$ or $\theta_2$.
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