Some hints. Let $c\colon I\to\mathbb R^2$ be immersive and $\alpha$ such that $\det(c'(\alpha),c''(\alpha))\neq0$. Define $J(x,y):=(-y,x)$.
Show that there is an open interval $ U\subset I$ containing $\alpha$ such that $\det(c'(a),c''(b))\neq0$ for all $a$, $b\in U$.
Let $r$, $s$, and $t\in U$ (with $r<s<t$).
The key point is that there exist $a$ and $b\in U$ such that (invoke the alternating mean value theorem by Schwarz)
$$2\det(c(s)-c(r),c(t)-c(r))=\det(c'(a),c''(b))(s-r)(t-r)(t-s)\neq0.$$
From here $c(s)-c(r)$ and $c(t)-c(r)$ are linearly independent, hence $c(r)$, $c(s)$ and $c(t)$ aren't collinear. Thus there's exactly one circle $S(r,s,t)$ through these points. Let $\rho(r,s,t)$ be its radius and $m(r,s,t)$ its center.
Calculate
$$m(r,s,t)=\frac{c(r)+c(t)}{2}+\frac{\langle c(s)-c(r),c(t)-c(s)\rangle}{2\det(c(s)-c(r),c(t)-c(r))}J(c(t)-c(r)).$$
Expand the second term by $1/(s-r)(t-r)(t-s)$. For $r,s,t\to\alpha$ it will converge to
$$\frac{\|c'(\alpha)\|^2}{\det(c'(\alpha),c''(\alpha))}Jc'(\alpha).$$
Finally $\rho(r,s,t)=\|c(r)-m(r,s,t)\|$.
EDIT The Schwarz Mean Value Theorem
Schwarz stated in the cited paper (in a more "modern" language and a bit generalised) the following:
Let $n$ and $m$ be natural numbers with $n\leq m$, $I$ a (non-empty) interval and $c\colon I\to\mathbb R^m=:V$ (or any other $m$-dimensional real vector space $V$) a $C^n$-path. Let $\omega\colon V^{n}\to\mathbb R$ be an alternating multilinear form. Let $t_0,\dots t_n\in I$ satisfying $t_0<\dots <t_n$.
Then there exist $x_i\in[t_0,t_i]$ for $i=0,\dots n$ satisfying $x_0\leq\dots\leq x_n$ such that
$$
\begin{align}
&\frac{\omega\bigl(c(t_1)-c(t_0),\dots, c(t_n)-c(t_0)\bigr)}{\prod_{0\leq i<j\leq n}(t_j-t_i)}=\\
&\quad\omega\left(\frac{c'(x_0)}{1!},\dots,\frac{c^{(n)}(x_n)}{n!}\right).
\end{align}
$$
(For $m=1$ and $\omega=\det$ this is the "usual" MVT.)
Best Answer
One definition of the curvature of a plane curve at a given point is $\tfrac{1}{\rho}$ where $\rho$ is the radius of the osculating circle to the curve at that point.
Consider a smooth curve in the plane, say $C$, and a fixed point on that curve, say $p$. There are lots of circles tangent to $C$ at $p$. In fact, you find that all of these circle have one important thing in common: all of their centres lie on the normal line to $C$ at $p$. (If you draw the tangent line to $C$ at $p$ then the normal line is the line through $p$ that is perpendicular to this tangent line.)
One of these tangent circles is different to all of the rest: it has higher order contact with $C$ at $p$. All but one of the circles are simply tangent to $C$. They meet $C$ like the the parabola $y=x^2$ meets the $x$-axis at the origin. As far as the first derivative is concerned, the circle and the curve are the same at $p$.
A single circle, called the osculating circle, has higher order contact with $C$ at $p$. Ordinarily, it meets $C$ like $y=x^3$ meets the $x$-axis at the origin. As far as the first two derivatives are concerned, the circle and the curve are the same at $p$.
Although, if $p$ is a vertex ($\kappa \neq 0$ and $\kappa' = 0$) then it meets $C$ like $y=x^4$ meets the $x$-axis at the origin. As far as the first three derivatives are concerned, the circle and the curve are the same at $p$.
(When I say "meets like", I mean up to diffeomorphism, i.e. up to smooth changes of coordinates.)
The radius of this osculating circle, $\rho$, is called the radius of curvature of $C$ at $p$ and $\tfrac{1}{\rho}$ turns out to be $\kappa.$ Interestingly, if $p$ is an ordinary inflection then $\kappa = 0$ (and $\kappa' \neq 0$) and so $\rho = \infty$. The osculating circle is centred at infinity and the circle becomes a line, i.e. the curve meets it tangent line line $y=x^3$ meets the $x$-axis at the origin.
The beauty of working with the osculating circles and not the horrible formula for $\kappa$ is that they are totally independent of coordinates. There is something natural and uncontrived about the contact between circles and curves. Moreover, circles and lines are the orbits of Euclidean transformations.