What is the intuition or geometric interpretation of $T_pS=\text{ker }df_p$? A regular surface $S$ given by the inverse function of a regular value of a function $a$, such as $S=f^{-1}(a)$ has the property that the tangent space to $S$ at a point $p$ is the kernel of the differential of f, i.e. $df_p.$
Without invoking the gradient of $f$ at $p,$ can I get a geometric intuition with an example of why this is true?
 A: As I said in the comments, if a curve, lies on S, post composing by f is a constant function (a), so its derivative is 0. By dimension counting, these vector spaces are equal.
Let's do the example $f(x,y,z)=x^2+y^2+z^2$. Then $S=S^2=f^{-1}(1)$ is the unit sphere, because $1$ is a regular value of $f$: 
$$ df = 2xdx+2ydy+2zdz ,$$ and at leasy one of $x,y,z$ is nonzero if $f(x,y,z)=1$. 
Let $c:[-1,1] \to S$ be any curve on the sphere $S$. Then $f\circ c(t) = 1$ for all $t$, so $(f\circ c)'(t)=0$ for all $t$, in particular for $t=0$. So $df(c'(0))=0$, and $c'(0) \in \ker_{c(0)}df$. Every tangent vector at $c(0)=p$ is the derivative of some curve on $S$, so we have $T_pS \subset \ker df_p$. But these are two vector spaces of dimension $2$, so they are equal (the same argument works for a hypersurface in $\mathbb{R}^n$, but I take it you are mainly interested in surfaces in $\mathbb{R}^3$).
A: If you have a curve $c:[-1,1]\rightarrow S$ such that $c(0)=p, f(c(t))=a$ and $df_p\circ c'(0)=0$, $c'(0)=lim_{t\rightarrow 0}{{c(t)-c(0)}\over t}$ if $S\subset\mathbb{R}^n$.
