A subset $S$ of a manifold $N$ of dimension $n$ is a regular submanifold of dimension $k$ if for every $p \in S$ there is a coordinate neighbourhood $(U,\phi)$ of $p$ in the maximal atlas of $N$ such that $U \cap S$ is defined by the vanishing of $n-k$ of the coordinate functions.
This is Tu's definition, An introduction to Manifolds, 2nd ed. Why do we demand equality between the set $U \cap S$ and the zero set of the vanishing coordinates? Tu stresses this fact in the example on the following page, where he excludes the chart corresponding to $V=(-2,0) \times (-1,1)$ in relation to the regular submanifold $S=(-1,1) \subset \mathbb{R}^2$. The inverse image of such a chart would include some points in the zero set outside of $U \cap S$, but is this a problem? We seem to restrict to $U \cap S$ anyhow. Were some similar questions in the past.
Added comment: So the author continues by introducing the restriction of the first $k$ components of a chart $(U,\phi)=(U,x^1,…,x^n)$ to $U \cap S$, $\phi_S: U \cap S \to \mathbb{R}^k$, $\phi_S=(x^1,…,x^k$), and shows that if $\{(U,\phi)\}$ is a collection of charts with the desired properties then $\{(U \cap S, \phi_S)\}$ is an atlas for $S$. To do so, he shows that the corresponding transition functions are smooth, which seems to hold in either case. In a subsequent section he also shows that the map $g:N \to S$ induced by a smooth map $f: N \to M$, whose image $f(N)$ lies in a regular submanifold $S$ of dimension $s$, is again smooth. In the proof, he notes that for a suitable chart $(V,\phi)$, relative to $S$, about a point $f(p) \in M$, he can pick an open set about $p \in U$ with $f(U) \subset V \cap S$, such that $\phi(f(q))=(y^1(f(q)),…,y^s(f(q)),0,…,0)$. From this he concludes, what seems to be nothing but trivial and from here on a matter of definition, with the words "It follows that on $U$, $\phi_S \circ g=(y^1 \circ f,…,y^s \circ f)"$, sounding as if there were anything more to it. The latter also seems to be a recurring theme for a build up that initially seems to be a matter of finding suitable definitions (excepting that such abstractions may be far from obvious) to go in accord with the comparatively deep results from Calculus and the Implicit function theorem.
Best Answer
Tu requires that for every $p \in S$ there exists a chart $(U,\phi)$ on $N$ with $p \in U$ such that $\phi^{-1}(\tilde{\mathbb R}^k) = U \cap S$. Here $\tilde{\mathbb R}^k = \{(x_1,\ldots,x_k,0,\ldots, 0 ) \mid x_1,\ldots,x_k \in \mathbb R \}$.
It seems that you ask why he doesn't relax this to the requirement that $U \cap S \subset \phi^{-1}(\tilde{\mathbb R}^k)$.
If we agree that $S$ should inherit from $N$ a natural structure of a smooth $k$-manifold, then this relaxed requirement causes indeed a problem.
As an example take $N = \mathbb R^2$. In the relaxed sense any subset $S$ of the $x$-axis would be a regular submanifold of dimension $1$. I think this is not a reasonable approach.
What we could do is to require that $U \cap S$ is an open subset of $\phi^{-1}(\tilde{\mathbb R}^k)$. This is the case in Tu's example in Fig. 9.1.
In fact, under this assumption there is an open subset $U^* \subset U$ such that $(U^*, \phi^* = \phi \mid_{U^*})$ is adapted chart relative to $S$, and this means that both definitions are equivalent.
How to find $U^*$? We know that $\tilde{\mathbb R}^k$ is closed in $\mathbb R^n$, thus $C = \phi^{-1}(\tilde{\mathbb R}^k)$ is a closed subset of $U$. Since $U \cap S$ is open in $C$, we see that $C^* = C \setminus (U \cap S)$ is closed in $C$, thus closed in $U$. Now take $U^* = U \setminus C^*$.