A subset $S \subseteq M$ is a regular submanifold of dimension $k$ of a smooth manifold $M$ of dimension $n$ if for every point $p \in S$, there exists a chart $(U_p, \varphi_p)$ around $p$ in the smooth structure of $M$ such that the intersection $U_p \cap S = \varphi_p'(U_p)$ where $\varphi_p'$ some function which may be obtained by sending $n - k$ of the coordinate functions of $\varphi_p$ to zero.
The above is supposed to be a definition, but I don't understand how it defines the topology or the smooth structure. Is the subspace topology implicit in the above definition? Or is it supposed to be another assumption?
Let $\varphi_S: U_p \cap S \to \mathbb{R}^k$. How can we show $(U_p \cap S, \varphi_S)$ is a chart on $S$? Does this process of constructing charts uniquely determine the smooth structure?
Yes, $S$ is given the subspace topology inherited from $M$. But the phrase
does not make much sense. The chart $\varphi_p : U_p \to V_p$ is a homeomorphism onto an open $V_p \subset \mathbb R^n$. Now the set $U_p$ occurs on both sides of the equation $U_p \cap S = \varphi_p'(U_p)$ and therefore $\varphi_p'$ must be a map on $U_p$ and taking values in $M$ - but this has no coordinate functions which can take the value $0$. What is meant is that $\varphi_p(U_p \cap S) = V_p \cap \mathbb R^n_k$, where $\mathbb R^n_k = \{(x_1,\dots,x_n) \in \mathbb R^n \mid x_{k+1} = \dots = x_n = 0\}$. Then you can define $$\varphi'_p= \pi^n_k \circ \varphi_p \mid_{U_p \cap S}: U'_p = U_p \cap S \to V'_p $$ where $\pi^n_k$ denotes projection onto the first $k$ coordinates and $V'_p = \pi^ n _k(V_p \cap \mathbb R^n_k)$.
Then you can prove that the $\varphi'_p$ form an atlas on $S$ which has smooth transition functions and thus determines a smooth structure on $S$.