TITLE: Why are two notions of Gaussian curvature are the same - what is the simplest & most didactic proof?
QUESTION [23 upvotes]: This question is still wide open - all of the answers so far rely on magical calculations. I've only accepted an answer because, by bounty rules, otherwise one would be accepted automatically. I can't change the accepted answer, but it would be amazing to have more discussion on this question.
I'd like a nice proof (or a convincing demonstration), for a surface in $\mathbb R^3$, that explains why the following notions are equivalent:
1) Curvature, as defined by the area of the sphere that Gauss map traces out on a region.
1.5) The integral of the product of principal curvatures.
2) The angle defect of parallel transport about a geodesic triangle.
(This equivalence may be considered as either a part of the Theorema Egregium or a part of Gauss-Bonnet. Proving that numbers 1 and 1.5 are the same is pretty easy).
Motivation: I'm teaching a five-day class for very bright high-school students. The idea is to give them an impression of what geometry is about. However, when I looked at Spivak's proof of this, it was much more of a messy calculation than I expected. I'd like, if at all possible, something more conceptual, ideally with a nice picture attached to it.
Since this doesn't have to be a perfectly complete class, I'll be perfectly happy with a good illustration of why this is true instead of a rigorous proof, if a conceptual and rigorous proof is completely out of the question. 
One idea I had is to show the example of a sphere and the hyperbolic plane, and then explain that on very small scales the curvature is constant. However, then I would need a nice proof that the embeddings of the hyperbolic plane in $\mathbb R^3$ have curvature -1.
Thank you very much!
P.S. This question is related, but not quite the same (I hope), to this question:
Equivalent definitions of Gaussian curvature
P.P.S. Thank you to whomever recommended to Berger's "Panoramic View of Riemannian Geometry". it was quite useful to me. I do not know why you deleted your answer. 
That books claims there is no conceptual proof. However, I'd still be very happy with a nice illustration of why one should believe this, especially for negative curvature.

REPLY [4 votes]: There is a physical "proof" of this fact which I learned from Mark Levy;
it is in his book "THE MATHEMATICAL MECHANIC: Using Physical Reasoning to Solve Problems".
Imagine that you keep the axis of a bicycle wheel
and move it in such a way that the bicycle wheel lies in
the tangent plane to the surface.
In the initial position the wheel stays still; 
you go along the loop in the surface and stop. After that your wheel rotates by some angle $\alpha$.
If your loop was triangular this angle is its defect; this is (2) in your list. 
The parallel motion does not rotates the wheel, 
so the same result will be the same if you only rotate the axis without moving the center of the wheel.
This tells you that $\alpha$ depends only on the spherical image of the loop
and from here it is easy to see that it is proportional to the algebraic area of the domain bounded by the spherical image of the loop.
I.e., the area of the sphere that Gauss map traces out on a region;
this is (1) in your list.