= Solution
For a unit-speed space curve, set
$$
t=\dot\alpha,\qquad
\kappa=|\dot t|,\qquad
n=\frac{\dot t}{\kappa},\qquad b=t\times n.
$$
These form the Frenet trihedron. With the signed torsion convention compatible with the question, the <Frenet-Serret formulas> are
$$
\dot t=\kappa n,\qquad
\dot n=-\kappa t-\tau b,\qquad
\dot b=\tau n.
$$
For an oriented surface with <unit normal> $N$, the signed <geodesic curvature> is
$$
\kappa_g=\langle\dot t,N\times t\rangle.
$$
On the unit sphere $N=\alpha$. Since $\alpha\cdot t=0$, <differentiation> gives
$$
1+\kappa\,\alpha\cdot n=0.
$$
Write $\alpha=An+Bb$. Then $A=-1/\kappa$, and differentiating this expression and comparing the $n$ coefficient with $\dot\alpha=t$ gives
$$
\dot A+\tau B=0,\qquad
B=-\frac{\dot\kappa}{\tau\kappa^2}.
$$
Thus, wherever the displayed quantities are defined,
$$
\boxed{\alpha=-\kappa^{-1}n-\tau^{-1}\kappa^{-2}\dot\kappa\,b}.
$$
Finally,
$$
\kappa_g
=\kappa n\cdot(\alpha\times t)
=-\frac{\dot\kappa}{\tau\kappa},
$$
which is the stated <curvature decomposition for a spherical curve>.
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