= Solution
For the <axis-angle composition of orthogonal spin rotations>, set $A=\Omega T/2$ and $B=\Phi/2$. Multiplication, including $\sigma_z\sigma_x=i\sigma_y$, gives
$$
U_\Omega U_\Phi=C I-i\mathbf v\cdot\boldsymbol\sigma,
\quad C=\cos A\cos B,
\quad\mathbf v=(\cos A\sin B,\ \sin A\sin B,\ \sin A\cos B).
$$
One checks that $C^2+|\mathbf v|^2=1$. For $s=|\mathbf v|\ne0$, choose
$$
\boxed{\vartheta=2\operatorname{atan2}(s,C),\qquad
\hat{\mathbf n}=\frac1s(\cos A\sin B,\ \sin A\sin B,\ \sin A\cos B).}
$$
This choice retains the full <spinor> sign and gives $U_\Omega U_\Phi=e^{-i\vartheta\hat{\mathbf n}\cdot\mathbf S}$. Repeated periods have the same axis:
$$
\boxed{U(NT-0^+,-0^+)=e^{-iN\vartheta\hat{\mathbf n}\cdot\mathbf S}.}
$$
Thus the <spin> expectation <vector> rotates through $N\vartheta$, interpreted modulo $2\pi$ as a spatial <rotation>. The positive $y$ component comes from the stated ordering; reversing the two <rotations> changes that component's sign.
When $s=0$, the one-period operator is $+I$ or $-I$, and no unique <rotation> axis exists. The latter is a $2\pi$ spatial <rotation> with a minus sign on a spin-$1/2$ state. It is safer to retain this central group element than to replace the angle by zero in a <spinor> calculation.
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