$\def\RR{\mathbb{R}}$I finally managed to do the routine computation Ted Shifrin describes, and I'm writing up the details for the record.
It turns out that the Gauss map is a bit of a red herring. Rather, let $X$ and $Y$ be two surfaces in $\RR^3$ and let $f: \RR \to X$ and $g: \RR \to Y$ be two curves which have the following interesting property: For all $t$, the tangent planes $T_{f(t)}X$ and $T_{g(t)}Y$ are parallel. Let $\theta(t)$ be the angle between $f'(t)$ and $g'(t)$. Then $d \theta$ is the difference between the geodesic curvature $1$-forms. (This is a slightly vague statement that will be made better below.)
In particular, $Y$ could be $S^2$, and $g$ could be the composition of $f$ with the Gauss map $X \to S^2$, but this case isn't special in any way.
First, I need to explain how to write the geodesic curvature as a $1$-form.
This is surely well known to experts, but most of the books I've been looking at define it as a number, and I had to convert to $1$-forms before I understood what was going on.
Let $C \subset X \subset \RR^3$ be a curve in an oriented surface in $3$-space. For $x \in X$, let $n_x$ be the unit normal vector. We define a $1$-form $\kappa_{C/X}$ on $C$ as follows: Let $\phi: \RR \to C$ be a parametrization of $C$ (or a segment of $C$). Let $\phi'(t) = |\phi'(t)| u(t)$, so $u(t)$ is a unit vector. Write $u'(t)$ for the derivative of $u(t)$ with respect to $t$. We define $\kappa_{C/X}$ so that $\phi^{\ast} \kappa = \det(u(t)\ u'(t)\ n_{u(t)}) dt$. Clearly, if $\phi$ is the arc length parametrization, then $\kappa$ is $(\mbox{geodesic curvature}) d (\mbox{arc length})$. But I claim that this definition works for any parametrization.
The reason is as follows: Suppose we change to another parametrization, with parameter $s$. Then $u$ and $n$ don't change. We have $du/dt = (du/ds) (ds/dt)$. But $dt = (ds/dt)^{-1} ds$. Since $\det$ is linear, these cancel.
I also need
Lemma Let $a$, $b$ and $c$ be three vectors in $\RR^3$ such that $a$ is a unit vector and $b$ is normal to $a$. Then $(a \times b) \cdot (a \times c) = b \cdot c$.
Proof We may assume that $a = (0,0,1)$. Then $b = (b_1, b_2, 0)$ and $c=(c_1, c_2, c_3)$. The claimed result is $(b_2, -b_1,0) \cdot (c_2, -c_1,0) = (b_1, b_2, 0) \cdot (c_1,c_2,c_3)$, which is obvious. $\square$
Now, let $X$, $f$, $Y$, $g$ and $\theta$ be as above. We write $n(t)$ for the common normal to $T_{f(t)} X$ and $T_{g(t)} Y$. Write $f(t) = |f(t)| u(t)$ and $g(t) = |g(t)| v(t)$, so $u$ and $v$ are unit vectors.
From our previous computations about geodesic curvature, we want to show
$$\frac{d \theta(t)}{dt} = \det(v(t), v'(t), n(t)) - \det(u(t), u'(t), n(t))$$
where prime is differentiation with respect to $t$. (Note that it was important to work out the formula for geodesic curvature without assuming we have an arc length parametrization, since it is unlikely that both $f(t)$ and $g(t)$ are unit speed.)
Now, since $u$ and $v$ are unit vectors, $\theta = \cos^{-1}(u \cdot v)$. We have
$$\frac{d \theta}{dt} = - \ \frac{d (u \cdot v)/dt}{\sin \theta} = - \ \frac{u' \cdot v + u \cdot v'}{\sin \theta}$$
so we must show that
$$u' \cdot v + u \cdot v' = - \det(v, v', (\sin \theta) n) + \det(u, u', (\sin \theta) n). \quad (\ast)$$
But $u$ and $v$ are unit vectors and $n$ is normal to both of them, so $(\sin \theta) n = u \times v$ and we can write the right hand side of $(\ast)$ as
$$\det(v, v', v \times u) + \det(u, u', u \times v) = (v \times v') \cdot (v \times u) + (u \times u') \cdot (u \times v).$$
We must show that
$$u \cdot v' + u' \cdot v = (v \times v') \cdot (v \times u) + (u \times u') \cdot (u \times v).$$
The result now follows from the Lemma, using $(a,b,c) = (v, v', u)$ in the first case and $(u, u', v)$ in the second. (Since $u$ and $v$ are unit length, we have $u \perp u'$ and $v \perp v'$, as the lemma requires.) $\square$