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!
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.
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.
