In my handout it is said that a circle $\Gamma(s)$ that is tangent to second order to a curve $\xi:[a,b]\to \mathbb R^2$ with unit speed and with curvature $\kappa$, then the radius of $\Gamma(s)$ is $1\over |\kappa (s)|$.

I don't quite understand what "tangent to second order" means. Is there a simple/intuitive way to see this?

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"Tangent to second order" means that, at the point of intersection $p$, the circle and the curve not only agree in velocity, but also in acceleration. If you take a Taylor expansion of both $\Gamma$ and $\xi$ at $p$, the first two terms will be the same.
A straight line is a constant-velocity curve. What's a constant-turning curve? A circle. So to measure curvature, you ask, "If the track disappeared in an instant and the particle kept turning at the same rate it's turning now, how big a circle would it follow?" If it's a small circle, the particle has to turn very hard, so, high curvature. If it's a large circle, the particle doesn't have to turn very hard, so, low curvature. If it's a straight line, just think of it as a circle of infinite radius, with curvature "$1/\infty$"$=0$.