Let's look at what do Carmo actually writes for the definition of a regular surface:
A subset $S \subset \mathbb{R}^3$ is a regular surface if, for each $p \in S$, there exists a neighborhood $V \subseteq \mathbb{R}^3$ and a map ${\bf x} : U \to V \cap S$ of an open set $U \subseteq \mathbb{R}^2$ onto $V \cap S \subset \mathbb{R}^3$ such that:
${\bf x}$ is differentiable. This means that if we write
$${\bf x}(u, v) = (x(u,v), y(u,v), z(u,v)), \qquad u, v \in U$$
the functions $x, y, z$ have continuous partial derivatives of all orders in $U$.
${\bf x}$ is a homeomorphism. [...]
(The regularity condition) For each $q \in U$, the differential
$d{\bf x}_q$ is [injective].
The mapping $\bf x$ is a parametrization or a system of (local) coordinates in (a neighborhood of $p$. The neighborhood $V \cap S$ of $p$ in $S$ is called a coordinate neighborhood.
This is section 2-2, definition 1 in do Carmo.
Notice a couple of things. Since $V \cap S$ doesn't have a a smooth manifold structure yet, we can't strictly speaking talk about ${\bf x}$ being differentiable, and if you look at the definition do Carmo gives what he's really saying is that the composition
$$U \xrightarrow{\bf x} S \cap V \hookrightarrow \mathbb{R}^3$$
is differentiable. Similarly, when he talks about $d{\bf x}_q$, he's talking about the differential of this same map, since again we can't talk about ${\rm T}_q (S \cap V)$ yet as it hasn't been defined.
Okay, now a smooth embedding, according to Lee, is an injective immersion which is a homeomorphism onto its image. Since $i$ the inclusion of a subspace, the topological conditions are already satisfied, so it remains only to check that $i$ is an immersion, i.e. that $i_\ast : {\rm T}_P S \to {\rm T}_P \mathbb{R}^3$ is injective for each $P \in S$.
Take a local parametrization ${\bf x} : U \to S$ at $P$, say with ${\bf x}(0) = P$. By definition, the composition
$$U \xrightarrow{\bf x} S \xrightarrow{i} \mathbb{R}^3$$
is an immersion. Since $(i \circ {\bf x})_\ast = i_\ast \circ {\bf x}_\ast$, the composition
$${\rm T}_0 U \xrightarrow{{\bf x}_\ast} {\rm T}_P S \xrightarrow{i_\ast} {\rm T}_P \mathbb{R}^3$$
is injective. Above, you've said you believe that $S$ is a 2-manifold. Given this, $\dim {\rm T}_0 U = \dim {\rm T}_P S = 2$ and $\dim {\rm T}_P \mathbb{R}^3 = 3$. So by elementary linear algebra the only way for the composition to be injective is for ${\bf x}_\ast$ to be an isomorphism and $i_\ast$ to be injective.
I found this proof confusing, too. Here is my understanding: we have that $D$ is an embedded submanifold of $M.$ Take any $p\in \partial D$. Then, there is a chart $\textit{in M},\ $ say, $(V,(x^1,\cdots,x^n))$ about $p$ such that $(D\cap V,(x^1,\cdots,x^k))$ is a boundary (slice) chart for $D$ about $p$. Thus, $q\in D\cap V\Rightarrow x^k(q)\ge 0$ But $\text{dim}\ D=\text{dim}\ M\Rightarrow k=n$, and so $x^n\ge 0.$ Now, $M$ is a manifold without boundary, which means that there must be a point $q\in V$ such that $x^n(q)<0$, (because $(V,(x^1,\cdots,x^n))$ is a chart about $p$ in $M$), which in turn implies that $q\notin D$ (because $D\cap V$ has all $x^n\ge 0$). Therefore, $V$ contains points in $D$ and in $M\setminus D$.
Best Answer
We are in $\mathbb{R}^n$ all the way, so the computations are more concrete.
To see why $(1) \implies (2)$, we can use the fact that we have a very explicit derivative for $\Phi$:
$$\Phi'_p=\begin{pmatrix} \nabla \Phi_1 \\ \nabla\Phi_2 \\ \cdots \\ \nabla \Phi_k \end{pmatrix}. $$
Since we are supposing $S$ is a regular level set, we have that those $\nabla \Phi_i$ are all linearly independent along $S$. They are also all normal to $S$, since $\Phi$ is constant there. This gives a global framing of the normal bundle.
To see why $(2) \implies (1)$, you simply use the fact that by assumption there exists a diffeomorphism $\Psi: NS \to S \times \mathbb{R}^k$, and consider $\Phi:=\pi_2 \circ \Psi \circ T,$ where $T: U \to V \subset NS $ is a diffeomorphism of a neighbourhood of $S$ onto a neighbourhood of the zero section on the normal bundle (such diffeomorphism is given by the tubular neighbourhood theorem).