curvature
The sign of the Gaussian curvature can be used to classify points as elliptic, hyperbolic, and parabolic. Wikipedia has this image with example surfaces:
I see how a hyperboloid surface has hyperbolic points, and respectively a sphere or ellipsoid has elliptic points. So these are aptly named.
But why are points with zero Gaussian curvature named parabolic, not something like cylindrical or conical?
Especially, a paraboloid has positive curvature.
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.
Some lines before that statement they say what they mean by a 'normal section' and denote the curvature of the resulting line by 'normal curvature' of that normal section. What is true is that, in the hyperbolic case, there are two directions for which the corresponding normal curvatures are $=0$ (because one principal curvature is negative, one is positive, and the normal section varies continuously if you rotate the plane orthogonal to the surface around the normal to the surface). The use of the name 'sectional curvature' is incorrect here, but very likey deemed natural to the author since the curvature he has in mind is derived from intersecting the surface with a normal plane.
The sectional curvature is in fact something else and, as you said, coincides with the (intrinsic) Gaussian curvature in case of two dimensional surfaces.
Edit: see also here for asymptotic lines and directions.
Best Answer
Nomenclature is from origin of conics categorization . Among the conics eccentricity $\epsilon$ for a hyperbola, parabola and ellipse are $ \gt 1, = 1, \lt 1 $ respectively. In its categorizing work sign of double or Gauss curvature relates to $ (1- \epsilon). $
In the equation of conic( two dimensions) there is already an indicator of things to come when it would be embedded in 3-space. For $ a x^2 + 2 h x y + by^2 $ + linear terms =0, then the sign of invariant $ a \cdot b - h^2 $ also decides to which of the three types the conic under consideration belongs.
Accordingly in $ \mathbb R^{2}$ say for a surface in Monge form $ z= f(x,y), K= (r \cdot t - s^2)/(1+p^2+q^2)^2 $ (partial derivatives of z) decides sign of Gauss curvature K, i.e., to which of the three types you have shown the surface belongs.
EDIT 1:
Also if a reputed mathematician had called it that way and it stuck,the matter of distinguishing between the three was settled temporarily.
Having said the above in defense of what I believe is the staus quo ante I am in agreement with the OP about inappropriateness of the parabolic appellation of $ K=0 $ flat surfaces.
When a paraboloid is and quite apparently looks $K >0$ there is no need to cling on to any historical reasons. In line with the view of OP I too like to see it changed to developable or flat, but not further retain parabolic.