A slightly over-simplified way of looking at intrinsic curvature of a plane slice is this:
Draw a small triangle on the plane slice. (How do you draw a triangle? Pick three points, and draw or compute the curve of shortest distance between each pair of vertices).
Compute the angular excess (the sum of the angles of the triangles) minus 180 degrees, and divide it by the area of the triangle. This is the intrinsic curvature of the plane slice.
If you do this on a flat sheet of paper, you'll get zero. If you do this on the surface of a sphere, you'll get a positive number. This might not be obvious at first glance, a bit of reading might be in order. I'll give one example to demonstrate, that example is a triangle on the surface of a sphere that has three right angles. You draw this triangle by starting at the north pole, going down to the equator, making a 1/4 circuit around the equator, an going back up to the north pole.
(On a sphere, the curves of shortest distance between any two points are great circles. ALl the curves we draw are arcs of great circles, so it satisfies our requirements that about the triangle being constructed by the curves of shortest length connecting the three points).
All the angles of this (spherical) triangle are right angles, and 3*90 = 270, so this triangle has an angular excess.
The area of this triangle is 1/8 the surface area of the sphere.
Thus we can quantify the fact that the surface of a sphere is curved, while a plane is flat, with this notion of intrinsic curvature, which we can compute by actual measurements on the surface.
As far as I know, there aren't any hidden gotchas (as long as you use _small_ triangles), and I think the above is equivalent to the more formal defitions that involve "parallel transport", while avoiding this unfamiliar term which would probalby require at least a post (and more preferably a book) to explain. But I don't at this point have a rigorous proof that this simplified definition is completely equivalent to the more formal one.
I think I got the sign reight too (a sphere has a positive curvature).
Note that if you have a space of higher dimension than 2, you'll have a different intrinsic curvature for every different plane slice. It turns out there is a rank 4 tensor that will give you the curvature of all possible plane slices and that this rank 4 tensor defines the general notion of intrinsic curvature in a space of dimension higher than 2. This may be too advanced to get into more details at the moment, it ight be better to stick with the above notion of curvature of a plane slice, which is known as "sectional curvature".
. What you are seeing is a geometrical shape embedded in a higher dimensional space and seeing how this shape bends in that higher dimensional space. Going back to the example of the graph of the function ##f(x) = x^{2}## versus the graph of the function ##f(x) = -x^{2}##, we visualize the two as an upwards facing bowl and downwards facing bowl, respectively, as embedded in two dimensional euclidean space. Their extrinsic curvatures then differ by a sign because of the flip in orientation. 
