EDIT, Thursday evening: a good illustration for the bit about negative curvature is the inner circle in the "hole" in a torus of revolution, which is a "closed geodesic" and cannot be contracted continuously to a point. The closed geodesic occurs in the (inner) part of the torus where the Gauss curvature is negative, while the outer portion has positive curvature. Meanwhile, no compact surface in $\mathbb R^3$ can have negative Gauss curvature everywhere, for which Neal gave a simple answer in comments. It is, of course, possible to have noncompact surfaces with negative curvature, such as the catenoid of revolution. However, for any such surface of revolution with negative curvature, the absolute value of the curvature goes to $0$ as we get far from the central axis. So Mariano asked whether there are any infinite surfaces in $\mathbb R^3$ with Gauss curvature that is bounded away from $0,$ and the answer is no, not for a $C^2$ surface. Link to the question MARIANO QUESTION.
ORIGINAL: This is Gauss-Bonnet for polygons on an oriented surface. The familiar case should be this: the area of a geodesic triangle on the unit sphere is $\alpha + \beta + \gamma - \pi,$ where $\alpha, \beta, \gamma$ are the angles at the vertices of the triangle. Next, we define the external angles $\theta_1 = \pi - \alpha, \; \theta_2 = \pi - \beta, \; \theta_3 = \pi - \gamma, $ according to Figure 4-25 of do Carmo:
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The unit sphere has curvature $K=1,$ so the integral of $K$ over a polygon is just its area. Furthermore, the boundary arcs are geodesics, with geodesic curvature $k_g = 0.$ So the statement about this area is equivalent to
$$ \int \int_T \; K d \sigma \; + \theta_1 + \theta_2 + \theta_3 = 2 \pi. $$
Compare this with Gauss-Bonnet, formula (1) here:
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You may or may not be told this: in the simply connected hyperbolic plane of constant curvature $-1,$ the area of a triangle is $ \pi - (\alpha + \beta + \gamma),$ where $\alpha, \beta, \gamma$ are the angles at the vertices of the triangle. However, because $K = -1,$ this is once again the same as
$$ \int \int_T \; K d \sigma \; + \theta_1 + \theta_2 + \theta_3 = 2 \pi. $$
For your exercise, you are asked about an (oriented) "diangle" $D,$ with angles $0 \leq \alpha, \beta \leq \pi.$ If an angle is equal to $0,$ that is what the author calls a "cusp," see Figure 4-26, the right half. We take the external angles as before, and get
$$ \int \int_D \; K d \sigma \; + \theta_1 + \theta_2 = 2 \pi, $$
$$ \int \int_D \; K d \sigma \; + (\pi - \alpha) + (\pi - \beta) = 2 \pi, $$
$$ \int \int_D \; K d \sigma \; = \alpha + \beta. $$
What does this mean? We have $K < 0.$ The right hand side of the equation is nonnegative. If there is any interior to the "diangle," the integral is strictly negative. The only legal possibility is that the two geodesics are identical, both vertices are cusps, the two geodesic arcs are exactly the same, once out, once back.
Otherwise, and here is where the topology comes in, the two geodesic arcs do not bound a piece of surface as I was assuming, and the loop cannot be deformed homotopically to a single point, thus the surface is not simply connected.
Well, I hope that works for you. There is quite a bit of detail that could be added about orientation. Furthermore, I do not think a closed surface in $\mathbb R^3$ can have negative Gauss curvature. I know that a closed surface in $\mathbb R^3$ cannot have constant negative Gauss curvature.
When you join up the edges of the flat square to make a projective plane, what happens to its corners? Each corner gets identified with the opposite corner.
So at two points in the projective plane you have two corner bits of squares
identified. They may be flat, but round the corner, you have only $\pi$
worth of angle instead of $2\pi$. This means you can't extend the flat metric
to these corners, so the quotient isn't naturally a Riemannian manifold.
An easier illustration: consider the surface of a standard cube in $\Bbb R^3$.
Each face is flat. What about the edges? They aren't trouble, one can think
of each pair of adjacent faces as a $2\times1$ rectangle folded over, and
that has a flat metric. So if you remove the vertices, the surface of the
cube has a flat Riemannian metric all over it.
But those pesky vertices! If you make a little circuit about one of these,
in effect you're going through an angle of $3\pi/2$. If you parallel-transport a vector round the path, it come back turned through a right angle. This means
there's no way of extending this flat Riemannian metric to the vertex.
This works for every compact surface; you can express it as a simplicial complex,
and if you delete the vertices then you can put a flat metric on it. Alas
one can rarely extend it to the whole surface (certainly if it has nonzero
Euler characteristic).
Best Answer
Same idea, just a bit different perspective:
Consider a closed simple geodesic on a closed surface homeomorphic to the sphere. It may not cut the surface into two parts of equal area, but if we consider the weighted area ( multiply the area form by $K$) then the two parts have equal "area". Indeed, apply Gauss-Bonet to both pieces, that have a common boundary ( a geodesic).
Now, assume moreover that $K>0$. Then if the two geodesic would not intersect, while each cut the surface into equal parts, we would get the "area" between the curves to be $0$, contradiction.
Obs: this would be the same with Gauss-Bonet for the region, but it's not intuitive. The intuition: a closed simple geodesic cuts the surface into two "equal" parts.
$\bf{Added:}$ Let's explain why the Euler characteristic is $0$. We need to divide the region which is topologically the lateral surface of a cylinder into several polygonal regions. Convenient choices can be made. Then count $$\chi(R) = v - e + f$$ where $v$ is the number of vertices, $e$ the number of edges, and $f$ the number of faces. So cut the region $R$ into two rectangles by drawing two curves joining the top and bottom geodesics ( these curves do not need to be geodesic, just to be simple, and without intersections). We have $v = 4$ ( two on each geodesic), $e= 2 + 2 + 2$ ( $2$ on each geodesic, and the $2$ shunts), and $f=2$. Now we see $$\chi(R)= 4 - 6 + 2 = 0$$