Trivial normal bundle $NS$ equivalence

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).