Note: I worked through this with a slightly different convention on torsion (which I believe is more standard) before I realized the definition given above. This should account for the minus sign.
The standard Frenet-Serret Formula for a curve parametrized by arc length are
\begin{array}
\vec{T}^{\prime} =& \kappa(s) \vec{N}\\
\vec{N}^{\prime} =& -\kappa(s)\vec{T} + \tau(s) \vec{B} \\
\vec{B}^{\prime} =& -\tau(s)\vec{N},\\
\end{array}
where the primes indicate derivatives with respect to arc length and $\vec{T} = \alpha^{\prime}$, $\vec{N} = \frac{\vec{T}^\prime}{\lvert \lvert \vec{T}^{\prime}\rvert \rvert}$ and $\vec{B} = \vec{T} \times \vec{N}$.
Note that the ultimate goal is to express torsion in terms of the parametrization. With this in mind, we will start with an expression that expresses $\vec{B}$ in terms of the parametrization, differentiate and apply the Frenet-Serret equations.
Starting with the identity $\alpha^{\prime} \times \alpha^{\prime\prime} = \vec{T} \times \vec{T}^{\prime} = \kappa \vec{B}$ (this streamlines the differentiation slightly) and differentiating with respect to $s$ we obtain
$$ \alpha^{\prime} \times {\alpha}^{\prime\prime\prime} = \kappa^{\prime}\vec{B} + \kappa\vec{B}^{\prime}.$$
Using the Frenet-Serret formula, substitution yields the following equation
$$\alpha^{\prime} \times \alpha^{\prime\prime\prime} = \kappa^{\prime}\vec{B} - \kappa\tau \vec{N}.$$
"Dotting" both sides of the above equation with $\vec{N}$ and using the defining relations of $\vec{T}$, $\vec{N}$ and $\vec{B}$ and we obtain
$$ \left(\alpha^{\prime} \times \alpha^{\prime\prime\prime}\right)\cdot \vec{N} = -\kappa \tau,$$
which yields
$$ -\frac{1}{\kappa^2}\left(\alpha^{\prime} \times \alpha^{\prime\prime\prime}\right)\cdot \alpha^{\prime\prime} = \tau.$$
Applying the scalar triple product to the above gives
$$ \frac{1}{\kappa^2}\left(\alpha^{\prime} \times \alpha^{\prime\prime}\right)\cdot \alpha^{\prime\prime\prime} = \tau.$$
$\newcommand{\R}{\mathbb{R}}$
My preferred approach is to derive the Frenet-Serret equations from scratch without assuming arclength parameterization. The equations you want are then an easy consequence.
Let $\alpha: I \rightarrow \R^3$ be a smooth curve such that $\alpha'(t), \alpha''(t)$ are linearly independent for all $t \in I$. For each $t$, $\alpha'(t)$ is tangent to $\alpha$ and therefore,
$$
T = \frac{\alpha'}{|\alpha'|}
$$
is a unit tangent vector that points in the direction of the curve.
If we view $\alpha'$ as velocity, then
$$
\sigma = |\alpha'|
$$
is the speed. Therefore, the length of the curve is
$$
\ell = \int_{t\in I} \sigma(t)\,dt.
$$
Since
$$
\alpha' = \sigma T,
$$
the acceleration of $\alpha$ is given by
\begin{align}
\alpha'' &= \sigma' T + \sigma T'.
\end{align}
The assumption that $\alpha', \alpha''$ are linearly independent implies that $T'\ne 0$.
Since $T\cdot T = 1$,
$$
0 = (T\cdot T)' = 2T\cdot T'.
$$
It follows that the equation above is an orthogonal decomposition of $\alpha'$. In particular, it decomposes $\alpha''$ into a tangential term whose magnitude is how quickly the speed is changing and a normal vector that indicates how the directionof $\alpha$ is changing. In particular, if we let
$$
N = \frac{T'}{|T'|},
$$
then
$$
T' = \sigma\kappa N,
$$
where $\sigma\kappa$ measures how quickly the direction of $\alpha$ is changing and $N$ indicates the direction of the change in the direction of $\alpha$. Finally, there is a unique unit vector $B$ such that $(T, N, B)$ is an oriented basis of $\R^3$. We can now derive the equations satisfied by these vectors:
\begin{align*}
0 &= (T\cdot T)' = (N\cdot N)' = (B\cdot B)'\\
\implies\ T'\cdot T &= 0\\
0 &= (T\cdot N)'\\
&= T'\cdot N + T\cdot N'\\
&= \sigma\kappa N\cdot N + T\cdot N'\\
\implies\ N'\cdot T &= -\sigma\kappa\\
0 &= (N\cdot B)'\\
&= N'\cdot B + N\cdot B'\\
\implies\ B'\cdot N &= -N'\cdot B.\\
0 &= (T\cdot B)'\\
&= T'\cdot B + T\cdot B'\\
&= T\cdot B'\\
\implies\ B'\cdot T &= 0.
\end{align*}
Therefore, if we let
$$
N' = \sigma\tau B,
$$
it follows that
$$
\begin{bmatrix} T & N & B\end{bmatrix}'
=
\sigma
\begin{bmatrix} T & N & B\end{bmatrix}'
\begin{bmatrix} 0 & -\kappa & 0\\
\kappa & 0 & \tau\\
0 & -\tau & 0
\end{bmatrix}.
$$
These are the Frenet-Serret equations. Observe that if $c$ is parameterized by arclength, then they become the more familiar form of the equations.
Now observe that
\begin{align*}
\alpha'\times\alpha'' &= \sigma T\times (\sigma T + \kappa N)\\
&= \sigma \kappa T\times N\\
&= \sigma \kappa B\\
\implies\ B &= \frac{\alpha'\times \alpha''}{|\alpha'\times \alpha''|}\\
\text{and }|\alpha'\times \alpha''| &= \sigma\kappa\\
\alpha'\times (\alpha'\times \alpha'') &= \sigma T\times(\sigma\kappa B)\\
&= \sigma^2\kappa(T\times B)\\
&= \sigma^2\kappa N\\
&= |\alpha'||\alpha'\times\alpha''| N\\
\implies\ N &= \frac{\alpha'\times (\alpha'\times \alpha'')}{|\alpha'||\alpha'\times\alpha''|}.
\end{align*}
Best Answer
Follows my derivation of the curvature formula
$\kappa(t) = \dfrac{\vert \alpha''(t) \times \alpha'(t) \vert}{\vert \alpha'(t) \vert^3}, \tag 0$
which I believe to be reasonably canonical. I have not yet been able to derive the expression for torsion, though I have put considerable effort into my attempts. I will keep working on it and edit my solution, if I find one, into this answer.
The unit tangent vector $T(t)$ to the curve $\alpha(t)$ is clearly given by
$T(t) =\dfrac{\alpha'(t)}{\vert \alpha'(t) \vert} = \vert \alpha'(t) \vert^{-1}\alpha'(t), \tag 1$
where the magnitude of $\alpha'(t)$ is
$\vert \alpha'(t) \vert = \langle \alpha'(t), \alpha'(t) \rangle^{1/2}; \tag{2}$
we may find the rate of change of the magnitude of $\alpha'(t)$ along $\alpha(t)$:
$\vert \alpha'(t) \vert' = \dfrac{1}{2}\langle \alpha'(t), \alpha'(t) \rangle^{-1/2}(2\langle \alpha'(t), \alpha''(t) \rangle) = \langle \alpha'(t), \alpha'(t) \rangle^{-1/2}\langle \alpha'(t), \alpha''(t) \rangle$ $= \langle \langle \alpha'(t), \alpha'(t) \rangle^{-1/2}\alpha'(t), \alpha''(t) \rangle = \langle \vert \alpha'(t) \vert^{-1} \alpha'(t), \alpha''(t) \rangle = \langle T(t), \alpha''(t) \rangle. \tag{3}$
The speed or rate of change of arc-length $s$ along $\alpha(t)$ with respect to $t$ is
$\dfrac{ds}{dt} = \vert \alpha'(t) \vert, \tag 4$
from which
$\dfrac{dt}{ds} = \vert \alpha'(t) \vert^{-1} = \langle \alpha'(t), \alpha'(t) \rangle^{-1/2}. \tag 5$
We may also find the rate of change of the tangent vector $T(t)$ along $\alpha(t)$:
$\dfrac{dT(t)}{dt} = T'(t) = -\vert \alpha'(t) \vert^{-2} \vert \alpha'(t) \vert' \alpha'(t) + \vert \alpha'(t) \vert^{-1}\alpha''(t), \tag 6$
and in light of (3) this may be written
$\dfrac{dT(t)}{dt} = T'(t) = -\vert \alpha'(t) \vert^{-2} \langle T(t), \alpha''(t) \rangle \alpha'(t) + \vert \alpha'(t) \vert^{-1}\alpha''(t)$ $= -\vert \alpha'(t) \vert^{-1} \langle T(t), \alpha''(t) \rangle T(t) + \vert \alpha'(t) \vert^{-1}\alpha''(t) = \vert \alpha'(t) \vert^{-1}(\alpha''(t) - \langle T(t), \alpha''(t) \rangle T(t)). \tag 7$
We now introduce the first Frenet-Serret equation
$\dfrac{dT}{ds} = \kappa N; \tag{7.1}$
written in terms of the variable $t$ and the curve $\alpha(t)$ this becomes;
$\kappa(t) N(t) = \dfrac{dT(t)}{ds} = \dfrac{dT(t)}{dt}\dfrac{dt}{ds} = \vert \alpha'(t) \vert^{-1} \vert \alpha'(t) \vert^{-1}(\alpha''(t) - \langle T(t), \alpha''(t) \rangle T(t))$ $= \vert \alpha'(t) \vert^{-2}(\alpha''(t) - \langle T(t), \alpha''(t) \rangle T(t)); \tag 8$
since
$N(t) \cdot N(t) = T(t) \cdot T(t) = 1, \tag{8.1}$
we have
$\kappa^2(t) = \kappa(t) N(t) \cdot \kappa(t)N(t)$ $= \vert \alpha'(t) \vert^{-2}(\alpha''(t) - \langle T(t), \alpha''(t) \rangle T(t)) \cdot \vert \alpha'(t) \vert^{-2}(\alpha''(t) - \langle T(t), \alpha''(t) \rangle T(t))$ $= \vert \alpha'(t) \vert^{-4} (\alpha''(t) - \langle T(t), \alpha''(t) \rangle T(t)) \cdot (\alpha''(t) - \langle T(t), \alpha''(t) \rangle T(t))$ $= \vert \alpha'(t) \vert^{-4} (\langle \alpha''(t), \alpha''(t) \rangle - 2 \langle \alpha''(t), T(t) \rangle^2 + \langle \alpha''(t), T(t) \rangle^2)$ $= \vert \alpha'(t) \vert^{-4}(\langle \alpha''(t), \alpha''(t) \rangle - \langle \alpha''(t), T(t) \rangle^2). \tag 9$
We apply the vector identity
$\vert \mathbf A \times \mathbf B \vert^2 = \vert \mathbf A \vert^2 \vert \mathbf B \vert^2 - (\mathbf A \cdot \mathbf B)^2, \tag{10}$
where
$\mathbf A, \mathbf B \in \Bbb R^3, \tag{11}$
(cf. this wikepedia page) to the expression $\langle \alpha''(t), \alpha''(t) \rangle - \langle \alpha''(t), T(t) \rangle^2$,
taking
$\mathbf A = \alpha''(t), \tag{12}$
$\mathbf B = T(t); \tag{13}$
we immediately see that
$\langle \alpha''(t), \alpha''(t) \rangle - \langle \alpha''(t), T(t) \rangle^2 = \vert \alpha''(t) \times T(t) \vert^2; \tag{14}$
thus (9) becomes
$\kappa^2(t) = \vert \alpha'(t) \vert^{-4} \vert \alpha''(t) \times T(t) \vert^2, \tag{15}$
whence
$\kappa(t) = \vert \alpha'(t) \vert^{-2} \vert \alpha''(t) \times T(t) \vert. \tag{16}$
We now recall (1) and substitute $\vert \alpha'(t) \vert^{-1} \alpha'(t)$ for $T(t)$ on the right-hand side of this equation, and obtain
$\kappa(t) = \vert \alpha'(t) \vert^{-2} \vert \alpha''(t) \times \vert \alpha'(t) \vert^{-1}\alpha'(t) \vert = \vert \alpha'(t) \vert^{-3} \vert \alpha''(t) \times \alpha'(t) \vert$ $= \dfrac{\vert \alpha''(t) \times \alpha'(t) \vert}{\vert \alpha'(t) \vert^3}, \tag{17}$
the desired result.