Proof of Smale Theorem (Audin-Damian), Lemma 2.2.8 In Audin and Damian, p.43, there is a proof of the following lemma, relating pseudo-gradient vector fields adapted to $f$ on $V$, namely $X$ and approximation $X'$. Here, $\alpha_j$ a critical value of the Morse function $f$:

and in the course of the proof there is the claim of an embedding $\Psi$, in the bottom:

I believe there is a mistake in the image of this embedding, namely it should be from $D^k\times Q\times [0,m]$ to $f^{-1}([\alpha_j+\epsilon, \alpha_j+2\epsilon])$.
If it's as written, it doesn't make sense to claim that $\{0\}\times Q\times\{0\}$ is the embedding of $Q$ on $f^{-1}(\alpha_j+\epsilon)$ (same for $\{0\}\times Q\times\{m\}$).
But having fixed that, how do we know that $$\Psi_{\star}\left( -\dfrac{\partial}{\partial z} \right) = X\quad ?$$
In particular, I feel like to be able to get this pushforward to be exactly $X$, we would need to calibrate the $\epsilon$ accordingly, is that what is happening? We choose $\epsilon$ carefully and $\Psi$ carefully so that $\Psi$ stretches the tubular neighborhood onto $f^{-1}([\alpha_j+\epsilon, \alpha_j+2\epsilon])$ so that the unit speed vector field $-\dfrac{\partial}{\partial z}$ becomes $X$?
 A: I think I understood the issue. To guarantee that $-\dfrac{\partial}{\partial z}$ corresponds to $X$ we need an appropriate local chart so that $-\dfrac{\partial}{\partial z} = X = -\nabla f$ (in this part of the manifold we are looking at $X = -\nabla f$). We flow $Q\times D^k$ along the flow of $-\nabla f$.
More precisely, $\Psi$ is the result of flowing $Q\times D^k$ along $-\nabla f$ until we reach $f^{-1}(\alpha_j + \epsilon)$, which requires a time $s$.
And $s$ in fact can be computed explicitly. More on that later.
But for now, the embedding $\Psi:D^k\times Q\times [0,m]\to f^{-1}[\alpha_j+\epsilon, \alpha_j + 2\epsilon]$ exists because the flow of $X$ (or of any vector field) around regular points can always be represented as $-\dfrac{\partial}{\partial z}$ in local coordinates (Theorem 9.22 in Lee's Intro to Smooth Manifolds, for instance), and if it is so then the level sets have to be $R^{n-1}\times\{c\}, c\in \mathbb{R}$.
This makes evident that we can start at the tubular neighborhood of $Q$ which in this local coordinate system is $D^k\times Q\times \{d\}$, and go backwards along the $z$-axis to $D^k\times Q\times \{c\}$ at $f^{-1}(\alpha_j+\epsilon)$, and see $D^k\times Q\times [c, d]$ as an embedded product manifold in the local coordinates of $M$ we are at.
We immediately find the embedding $\widetilde{\Psi}:D^k\times Q\times [0,m]\to \widetilde{M}$, $m = d-c$, where we use $\widetilde{M}$ to indicate this map is to the local chart of $M$ where $X = -\dfrac{\partial}{\partial z}$.
Bringing this embedding out of the local coordinates we find the $\Psi$ we wanted.
Finding $s$.
To find $s$, we can use the Morse chart. We fix $q\in Q$. Then $f\varphi(q):\mathbb{R}\to \mathbb{R}$ is single variable function, with $\varphi$ the flow of $-\nabla f$. $$(\alpha_j + \epsilon) - (\alpha_j + 2\epsilon) = f\varphi_s(q) - f\varphi_0(q) = \int_0^s (f\varphi(q))'(t)dt = \int_0^s(\nabla f(\varphi_t(q)))\cdot \dfrac{d}{dt}\varphi(q)(t) dt$$
$$ = -\epsilon  = \int_0^s- \lVert \nabla f (\varphi_t(q))\rVert^2$$
Using the Morse chart where $f(x) = f(c_j) - x_1^2-...-x_\lambda^2+...+x_n^2$, the field is $-\nabla f = (2x_1,...,2x_\lambda,  ..., -2x_n)$, the flow from $(x_1, ..., x_n)$ is $(e^{2t}x_1, ..., e^{2t}x_\lambda, e^{-2t}x_{\lambda+1},..., e^{-2t}x_n)$, $$- \lVert \nabla f (\varphi_t(q))\rVert^2 = -4(e^{4t}\lVert x_\lambda\rVert^2 + e^{-4t}\lVert x_{n-\lambda} \rVert^2).$$
We conclude $$-\epsilon = \int_0^s  -4(e^{4t}\lVert x_\lambda\rVert^2 + e^{-4t}\lVert x_{n-\lambda} \rVert^2)dt = \Big\lvert_0^s -e^{4t}\lVert x_\lambda\rVert^2 + e^{-4t}\lVert x_{n-\lambda}\rVert^2$$
But we start with $q\in Q = f^{-1}(\alpha_j + 2\epsilon)\cap W^s(c_j)$, so that $- \lVert x_\lambda\rVert^2 + \lVert x_{n-\lambda} \rVert^2 = 2\epsilon$ and $\lVert x_\lambda\rVert^2 = 0$
so $$\epsilon = e^{-4s}\lVert x_{n-\lambda}\rVert^2 = e^{-4s}2\epsilon\Rightarrow \log 1/2 = -4s\Rightarrow s = -0.25\log(1/2).$$
