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.
I'm with you, although your sentence starting with "Clearly" could use an explicit proof. Namely, from $\mathbf t=\mathbf c$ constant, you should integrate to get $\boldsymbol\gamma(s)=s\mathbf c + \mathbf d$.
All I see on first principles from your lecturer's proof is that $\mathbf r$ lies in the plane through the origin with normal vector $\mathbf a$.
Best Answer
Given a straight line $\mathcal{D}$, any point $(x,y)$ of $\mathcal{D}$ must satisfy a relation as the following:
$$ ax + by = c.$$
Thus we have $$ ax' + by' = 0, \qquad ax'' + by'' = 0.$$ with $a, b$ not both zero.
Suppose that $a \neq 0$, then $$ 0 = ax'y'' + by'y'' = ax'y'' - ax''y',$$ which implies $$ 0 = x'y'' - x''y'.$$
So the numerator of the curvature is always $0$.