Some confusion about normal vector, curvature and normal curvature in Do Carmo's textbook. There is this part of a text from Do Carmo's Differential Geometry that I don't quite understand.

I understand the definitions of curvature, normal curvature, normal section etc. But what I am confused at is the part that says "In a neighbourhood of $p$, a normal section of $S$ at $p$ is a regular plane curve on $S$ whose normal vector $n$ at $p$ is $\pm N(p)$ or zero; its curvature is therefore equal to the absolute value of the normal curvature along $v$ at $p$."
First, is the neighbourhood of $p$ a part a curve $C$, a surface $S$ or a neighbourhood of $p$ on the normal section? It is not that clear. Especially when it mentioned the curvature it seems that it refers to the curve $C$. 
Another thing that confuses me is why is the normal vector either $N(p)$ or zero? And why is the curvature equal the absolute value of the normal curvature?
Here is how I worked it out: $k_n=k\cos\theta$ and $\theta=0$, so $k_n=k$, where $k_n$ is the normal curvature. Am I correct? Or should I reason as follows: $n=0$, so $\cos\theta=<0,N>=0$, so $k_n=0$?
I have been looking back and forth for the definitions and tried to work it out myself but still couldn't really understand the text. Could someone please help clarify the confusion? Thanks.
 A: "Neighborhood of $p$" refers to either a small region around $p$ on $S$, or a small regions around $p$ in 3-space -- either interpretation leads to the rest of the sentence. I'd choose the 3-space version. 
Part 2: suppose that your surface is a cylinder, and your "curve" is one of the straight lines parallel to the axis of the cylinder. Then the curve-normal would be zero. 
More generally: the plane chosen contains the surface normal $N$, right? That surface normal is perpendicular to the tangent plane of $S$ at $p$ (definition). And the tangent to any curve on $S$ that passes through $p$ is actually in the tangent plane to $S$ at $p$. (Indeed, some books define the tangent plane as the set of all tangents-to-curves-through-$p$.) So the tangent to the curve, which is a tangent to the surface as well, is perpendicular to $N$. And it lies in the slicing plane as well. 
Now look in the slicing plane: you've got a curve in there whose tangent IS a surface-tangent. The curve-normal, for a plane curve, is always perpendicular to the curve-tangent. So the curve-normal must be perpendicular to the curve-tangent, AND lie in the slice-plane. And there's only one possibility (up to sign/scale). Now look at N: it's both perpendicular and lies in the plane. So it satisfies the conditions. So it MUST be the curve-normal in the plane. 
