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You've basically walked right up to the explanation. The second-order approximation you describe, when $D = 0$, is

$$g(x, y) = a\left(x + \frac{b}{2a}y\right)^2$$

Define $z = x + \frac{b}{2a}y$, and we have $g(x, y) = g(z) = az^2$.

This means that $g$ describes a function that really only has one independent variable, a linear combination of $x$ and $y$. It looks like a parabola dragged along a line, and every point in the path of the vertex of the parabola is an extremum.

But since $g$ is only a second-order approximation to $f$, higher-order effects in $f$ may curve the path of the vertex with or against the curvature of the parabola, which will determine whether the critical point is an extremum or a saddle point. This isn't an issue when $D \neq 0$, because then the second-order effects will dominate the higher-order effects and knowing the behavior of $g$ near the critical point is sufficient.