How to derive these general formulas for the Frenet Frame of a curve not parameterized by arc-length? We may define the Frenet Frame $(T, N, B)$ of a regular curve $\alpha$ as follows:
$$T:=\frac{\alpha'}{| \alpha'|}$$
$$N:=\frac{T'}{|T'|}$$
$$B:=T\times N$$
and it can be shown that if $\beta$ is an arc-length parametrization of $\alpha$ by $s$, then $T_{\alpha}(t)=T_{\beta}(s(t))$,
$N_{\alpha}(t)=N_{\beta}(s(t))$, and
$B_{\alpha}(t)=B_{\beta}(s(t))$.

In this post both answers point to another way of expressing the Frenet frame of the curve $\alpha$, namely by defining $T$ as above and letting
$$N=\frac{\alpha'\times(\alpha''\times \alpha')}{|\alpha'|\ |\alpha'' \times \alpha'|}$$
$$B=\frac{\alpha'\times \alpha''}{|\alpha' \times \alpha''|}.$$

How can these last two formulas be proven equivalent to their former definitions?
 A: $\newcommand{\R}{\mathbb{R}}$
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*}
