Let $Y \subset \mathbb{R}^M$, then the normal bundle $N(Y)$ is a manifold. (page71)
First the proof defines $Y = \{ \psi (y) = 0 \}$ and take a open set $\tilde{U} \subset \mathbb{R}^M$ around a point $y$ and a submersion $\phi: \tilde{U} \to \mathbb{R}^{codim Y}$ with open set $\tilde{U} \cap Y = U = \phi^{-1}(0)$ in $Y$. Now they went on to prove $d\phi_y:\mathbb{R}^M \to \mathbb{R}^k$ has kernel $T_y(Y)$. I don't understand why the kernel is the tangent space.
So I tried rank nullity $\dim \ker d\phi_y + \dim Im d\phi_y = \dim \tilde{U}$
and I get $\dim \ker d\phi_y + codim(Y) = \dim \tilde{U}$. What is the dimension of $\tilde{U}$ here? And how do I know that $\dim \tilde{U} = M$?
Because if it is $M$ then $\ker (d\phi_y) \approx T_y(Y)$ (I am not sure why the isomorphism is identity though).
Short answer: $Y$ is locally defined as a level set of $\phi$, so d$\phi$ vanishes on a vector if and only if this vector is tangent to $Y$, (just like the differential of the "altitude" function is $0$ when you move along a level curve in the mountains).
With this in mind, to actually prove it with formulas, use the definition of the tangent space $T_yY$ as the set of all vectors that are derivatives at $t=0$ of paths $\gamma: [-1,1]\to \Bbb R^M$ such that $\gamma([-1,1])\subset Y$ and $\gamma(0)=y$, and use the chain rule when you compose $\gamma$ and $\phi$ to evaluate d$\phi$ on a tangent vector.
Note that the dimension of $\Bbb R^M$ and $\tilde{U}$ are the same because $\tilde{U}$ is open. But the fact that the dimensions are the same is too weak in the sense that yes, it does imply that there is an isomorphism as required, but there is no reason why this isomorphism would be natural in any way.