The theorem in question is as follows:
Let $V$ be a closed smooth manifold and $f: V\to \mathbb{R}$ a Morse function. Let $a$ be a critical point of $f$ with index $k$ and $\alpha=f(a)$. Suppose that for some $\epsilon>0$, $f^{-1}([\alpha-\epsilon, \alpha+\epsilon])$ is compact and does not contain any critical points distinct from $a$. For any sufficiently small $\epsilon>0$, the homotopy type of $V^{\alpha+\epsilon}=f^{-1}([\alpha+\epsilon,-\infty))$ is that of $V^{\alpha-\epsilon}$ with a cell of dimension $k$ attached.
In the beginning of the proof, we wish to construct a function $F$ which coincides with $f$ outside of a neighborhood of $a$ where $F<f$, such that $F^{-1}((-\infty, \alpha-\epsilon))$ is the union of $V^{\alpha-\epsilon}$ and a neighborhood of $a$.
This is done by taking a Morse chart $(U,h)$ of a neighborhood of $a$ and an $\epsilon>0$ such that $f^{-1}([\alpha-\epsilon,\alpha+\epsilon])$ is compact and $U$ contains a ball of radius $\sqrt{2\epsilon}$ with center $0$ and using a smooth function $\mu:\mathbb{R}^{\geq 0}\to \mathbb{R}^{\geq 0}$ such that \begin{align} \mu(0)&>\epsilon\\ \mu(s)&=0 \text{ for } s\geq 2\epsilon\\ -1<\mu'(s)&\leq 0 \text{ for every } s \end{align}
In the Morse chart $h: U\to V$ where $U\subset \mathbb{R}^{n}=\mathbb{R}^{k}\times \mathbb{R}^{n-k}$ open, $f$ takes the form $f(h(x_-, x_+))=\alpha-||x_-||^2+||x_+||^2$
Using this $\mu$, we can construct a function $$F(x)=\begin{cases} &f(x) &\text{ if } x\notin h(U)\\ &\alpha-||x_-||^2+||x_+||^2\mu(||x_-||^2+2||x_+||^2) &\text{ if } x=h(x_-,x_+) \end{cases}$$
The authors claim that outside of the set $||x_-||^2+2||x_+||^2\leq 2\epsilon$ one has $F=f$, but if $\mu(s)=0$ for $s\geq 2\epsilon$, then for $||x_-||^2+2||x_+||^2>2\epsilon$ we have $\mu(||x_-||^2+2||x_+||^2)=0$ and $$F(x)=\alpha-||x_-||^2\neq \alpha-||x_-||^2+||x_+||^2=f(x)$$ Is there a detail that I'm missing, or is there a typo?