What happens when the normal to surface is zero? In Differential Geom we're always given that surfaces should be regular, meaning the partial derivatives at every point are linearly independent, or the normal is non-zero. 
I get that the tangent space isn't well defined when the partial derivatives are linearly dependent. But I can't find any explanations as to what is happening to the surface itself. If someone could give a more geometrical/ intuitive reason why surfaces should be regular that would be great!
 A: Consider the "surface" defined by
$$
S(x, y) = (x^3, y, 1).
$$
Since the image is just the plane $z = 1$, it's clearly a nice surface, even though $\partial S/\partial x$ is zero at $x = 0$. So in this case your statement "the tangent space isn't well defined when the partials are dependent" is incorrect. It's a bit more subtle than that. But your statement's not TOO far wrong, and is a workable one to go with for now.
Then consider the following:
$$
S(x, y) = (x^3, |x^3|, y)
$$
This "surface" looks like an extruded letter "V", but its $x-$ and $y-$ partials are defined and smooth everywhere. It's more or less the last example, with the trouble with the zero-deriv being shown off rather than hidden. And that's why you need a well-defined normal.
A: Geometrically, the surface has a dent, or it degenerates to a line or curve, losing precisely the $2$-dimensionality that you want to study. For example, consider $$f(u) = \begin{cases} e^{-1/u} , &\text{if }u > 0 \\ 0, & \text{otherwise} \end{cases}$$
and plot using some program: $${\bf x}(u,v) = (f(u)\cos v, f(u)\sin v, u)$$
This is not regular because $f$ can be zero. Things will get bad in the $z$-axis, which is part of it.
