Curvature and Circumference of Circle Theorem Let $\gamma\colon [a,b]\rightarrow \mathbb{R}^2$ be a unit speed simple closed curve, with $\gamma'(a)=\gamma'(b)$ and $N$ is the inward-pointing normal.  Then
$$ \int_{a}^b \kappa_N(s)ds=2\pi. $$

Question: If a circle has curvature $\kappa$ and circumference $C$ then from above theorem, how does it follow that $\kappa C=2\pi$?
Ref: Riemannian Manifolds - John M. Lee, Theorem 1.6 and comments, p.4.
 A: I started preparing this answer before Narasimham made his edits, wherein the original independent variable name $t$ was replaced by $s$.  Since I was nearly done when his edits occurred, I chose to respect the original formulation in my answer; nevertheless, the relationship 'twixt $s$ and $t$ should be clear in what follows. 
A unit-speed circle in $\Bbb R^2$, centered at $(x_0, y_0)$ and of radius $\rho > 0$, may be parametrically represented as
$\gamma(t) = (x_0 + \rho \cos (\dfrac{t}{\rho}), y_0 + \rho \sin (\dfrac{t}{\rho})); \tag{1}$
taking $a = 0$ and $b = 2\pi$, we see that
$\gamma(a) = \gamma(0) = \gamma(2\pi) = \gamma(b).  \tag{2}$
Also,
$\gamma'(t) = (-\sin(\dfrac{t}{\rho}), \cos(\dfrac{t}{\rho})), \tag{3}$
whence
$\dfrac{ds}{dt} = \Vert \gamma'(t) \Vert = \sqrt{\sin^2(\dfrac{t}{\rho}) + \cos^2(\dfrac{t}{\rho})} = 1, \tag{4}$
where $s$ is the arc-length along $\gamma(t)$; from (4),
$s = t + s_0, \tag{5}$
which shows that both $s$ and $t$ effectively measure distance traversed along $\gamma(t)$, i.e. that $\gamma(t)$ is in fact a unit speed curve; the offset $s_0$ simply indicates a possible difference in parametrized  points on $\gamma(t)$ from which the measurement of distance is taken.  (By paremetrized point I mean a point $\gamma(t')$ on the curve together with a specific parameter value $t'$; this usage to disambiguate situations in which the same $\gamma(t')$ is in fact the image under $\gamma$ of more than one value $t'$, as in the present case with $\gamma(t' + 2\pi)  = \gamma(t')$ for all $t'$.)  If we agree, as we will, that $s$ is measured from $\gamma(0)$ at $t = 0$, then $s_0 = 0$ and $s$ and $t$ agree:
$s = t. \tag{6}$
We see from (4) that $\gamma'(t)$ is in fact the unit tangent vector field along $\gamma(t)$; in the notation of Frenet-Serret, we write
$T(t) = \gamma'(t); \tag{7}$
since $t = s$ we have the Frenet-Serret equation
$\dfrac{dT}{ds} = \dfrac{dT}{dt} = \kappa(t) N(t), \tag{8}$
where $\kappa(t)$ is the curvature, and $N(t)$ is the unit normal vector, to $\gamma(t)$ at the parameter value $t$; that is, at the point $\gamma(t)$ itself.  From (3), (7) and (8) we have
$\dfrac{dT}{dt} = \gamma''(t) = (-\dfrac{1}{\rho} \cos (\dfrac{t}{\rho}), -\dfrac{1}{\rho} \sin (\dfrac{t}{\rho})) = \dfrac{1}{\rho}(-\cos(\dfrac{t}{\rho}), -\sin (\dfrac{t}{\rho}))$
$= \dfrac{1}{\rho}N(t) = \kappa(t) N(t).  \tag{9}$
Inspecting (9), it is clear that (i.)
$\kappa(t) = \dfrac{1}{\rho} \tag{10}$\
is a constant; and (ii.)
$N(t) = (-\cos(\dfrac{t}{\rho}), -\sin (\dfrac{t}{\rho})) = -(\cos(\dfrac{t}{\rho}), \sin (\dfrac{t}{\rho})) = \dfrac{1}{\rho}((x_0, y_0) - \gamma(t)), \tag{11}$
which shows that $N(t)$ is directed from $\gamma(t)$ towards $(x_0, y_0)$, the center of the circle;  thus it is an inward-pointing normal vector.  Last but not least, we observe that
$\gamma'(a) = \gamma'(0) = \gamma'(2\pi) = \gamma'(b); \tag{12}$
we see that the criterion of the stated theorem are all met by the circle $\gamma(t)$; furthermore, since $\kappa(t) = \kappa(s) = 1/\rho$ is constant, 
we have
$\int_a^b \kappa(t) dt = \int_a^b \kappa(s) ds = \kappa \int_a^b ds = \kappa C; \tag{13}$
then by the theorem, the integral
$\int_a^b \kappa(t) dt = \int_a^b \kappa(s) ds = 2\pi; \tag{14}$
thus it follows that
$\kappa C = 2\pi, \tag{15}$
as was requested.
Nota Bene:  Establishing the formula (15) by calling on the much more general theorem given in the text of the problem seems a little bit like going after flies with a shotgun, since once we see that the curvature of the circle is a constant $\kappa = 1/\rho$, (15) follows immediately from the formula
$C = 2\pi \rho, \tag{16}$
with which I presume most readers are familiar; nevertheless, the relation (15) does provide perhaps the simplest possible case/application of the stated theorem, and so it is worthwhile to work it out; a collection of concrete, simple examples being in many ways the true grist which is ground by the mathematical mill.  End of Note.
