To rephrase the definition, for each $p\in S$, we have a neighborhood $U$ of $p$ and a coordinate chart $U\to V\subset \mathbb{R}^n$ of $N$ such that $S\cap U$ is the inverse image of a the intersection of $V$ with a $k$-dimensional linear subspace of $\mathbb{R}^n$ (in particular, a subspace of the form $(\star,\star,\star,\cdots, 0,0,0)$).
Locally, things do look like the $xy$-plane embedded in $\mathbb{R}^3$, defined by the vanishing of the $z$ coordinate. Of course, this won't hold for every chart, but we can cover $S$ with charts of $N$ such that, locally, a chart will determine what the points of $S$ are. If you have a point, and you have a chart, that tells you what the other points of $S$ are near $p$.
By the implicit function theorem, if $S$ is any submanifold of $N$, and $p\in S$, we can find a coordinate chart such that near $p$ the inclusion of $S$ into $N$ looks like the inclusion if $\mathbb{R}^k\subset \mathbb{R}^n$. Phrased differently, the condition that you wanted to hold happens for EVERY submanifold. So what this definition does is gives a topological restriction that you can't have different parts of the submanifold coming too close together.
For example, if you consider a slight modification of the topologist's sine curve, where you include a segment $(-1,1)$ of the $y$-axis and we connect that up with a path to the right hand side of the curve, you can view it as the image of the interval $(0,1)$ into $\mathbb{R}^2$ under a smooth, injective map, and so it is a submanifold of sorts. However, it is not a regular submanifold. Indeed, the induced topology (as a subset of $\mathbb{R}^2$ is different than the usual topology on $(-1,1)$). This is what the definition is meant to prevent, I believe.
Let $p = (0,\eta) \in I$. Assume $(U,\phi) = (U,x^1,x^2)$ is an adapted chart rel. $S$ with $p \in U$. Then $q \in U \cap S$ iff $x^2(q) = 0$.
$\phi$ is a diffeomorphism between $U$ and an open subset $V \subset \mathbb R^2$. In particular, the Jacobian determinant of $\phi$ at $p$ must be nonzero. Now let us compute $\frac{\partial x^2}{\partial x}(p)$ and $\frac{\partial x^2}{\partial y}(p)$.
Since $x^2$ vanishes on $I \cap U$, we get $\frac{\partial x^2}{\partial y}(p) = 0$.
There exists a sequence $p_n = (h_n,\eta) \in S$ such that $h_n > 0$ and $h_n \to 0$. Thus $p_n \to p$ which implies $p_n \in U$ for $n \ge n_0$. Then $p + h_n(1,0) = (0,\eta) + (h_n,0) = (h_n,\eta) = p_n$, hence
$$\frac{\partial x^2}{\partial x}(p) = \lim_{n \to \infty} \frac{x^2(p + h_n(1,0)) - x^2(p)}{h_n} = \lim_{n \to \infty} \frac{0 - 0}{h_n} = 0 .$$
This shows that the Jacobian determinant of $\phi$ at $p$ is zero, a contradiction.
Best Answer
First show that $\Phi_S(U \bigcap S) = \phi(U) \bigcap \mathbb R^k$, then use that $\phi(U) \subset \mathbb R^n$ is open, then use the subspace topology on $\mathbb R^k \subset \mathbb R^n$.