Flow reparametrization realizes a level isotopy

On $K$ the function $\lambda :=-df(X)$ is smooth and strictly positive, because $K$ has no critical point and $df(X)<0$ off the critical set; it is bounded below by a constant $c>0$ as $K$ is compact. Put $Z:=X\mathrm{/}\lambda$ on $K$, so $df(Z)=-1$. By [F1], equivalently reversing the normalized downward flow, $T(q,s)={\mathrm{\Phi }}_{b-s}^Z(q)$ is a level-preserving diffeomorphism $T:f^{-1}(b)\times [a,b]\to K$ with $f(T(q,s))=s$ and $T(q,b)=q$; in these coordinates $Z$ corresponds to $-{\mathrm{\partial }}_s$, and the $X$-trajectory transport $\phi :f^{-1}(a)\to f^{-1}(b)$ is $\phi (T(q,a))=q$.

Choose a smooth function $\alpha :[a,b]\to [0,1]$ with $\alpha =0$ on a neighbourhood of $a$ and $\alpha =1$ on a neighbourhood of $b$; it exists by [F3] after a translation and rescaling of the interval. Define a diffeomorphism $F$ of $f^{-1}(b)\times [a,b]$ by $F(q,s):=(h_{\alpha (s)}(q),s)$, with inverse $(q,s)\mapsto (h_{\alpha (s)}^{-1}(q),s)$; thus $F$ preserves the level coordinate, it is the identity near $s=a$, and its restriction to $s=b$ is $h_1$. Let $\widehat{X}$ be the field on $K$ whose coordinates under $T$ are $F_{\ast }(-{\mathrm{\partial }}_s)$.

The field $\widehat{X}$ is smooth, and $df(\widehat{X})=-1$: on each level set $f=s$ the differential of $F$ pushes $-{\mathrm{\partial }}_s$ to a vector whose level component is $-1$ and whose tangential component is horizontal, so $\widehat{X}$ descends the levels at unit speed. Because $\alpha$ is constant near the ends of $[a,b]$, $F_{\ast }(-{\mathrm{\partial }}_s)=-{\mathrm{\partial }}_s$ on a neighbourhood of $f^{-1}(a)\cup f^{-1}(b)$, hence $\widehat{X}=Z$ there. Define $X^{\mathrm{\prime }}:=\lambda \widehat{X}$ on the closed band $K$ and $X^{\mathrm{\prime }}:=X$ outside $K$. The two definitions agree on a neighbourhood of $\mathrm{\partial }K$, so $X^{\mathrm{\prime }}$ is a smooth field on all of $M$; on $K$ one has $df(X^{\mathrm{\prime }})=-\lambda <0$, and off $K$ the field is $X$, which is downward gradient-like and has all of its critical points outside $K$. Hence $X^{\mathrm{\prime }}$ is downward gradient-like for $f$ and equals $X$ outside $f^{-1}(a,b)$.

The field $X^{\mathrm{\prime }}$ is complete. A smooth trajectory confined to the compact band extends across every finite time endpoint by local flow existence and a finite chart cover. Moreover $f$ strictly decreases along every nonconstant $X^{\mathrm{\prime }}$-trajectory, so such a trajectory meets the band $K$ in at most one time interval; inside $K$ the coordinates $F^{-1}\circ T^{-1}$ turn $X^{\mathrm{\prime }}$ into a positive rescaling of $-{\mathrm{\partial }}_s$, with $\lambda \ge c>0$, so the time spent in $K$ is at most $(b-a)\mathrm{/}c$; outside $K$ trajectories are $X$-trajectories, and $X$ is complete. A maximal $X^{\mathrm{\prime }}$-trajectory therefore has no finite endpoint: after crossing $K$ it agrees with a maximal $X$-trajectory, which is defined on all of $\mathbb{R}$.

Compute the level transport of $X^{\mathrm{\prime }}$. Trajectories of $X^{\mathrm{\prime }}$ and of $\widehat{X}$ have the same images because $X^{\mathrm{\prime }}=\lambda \widehat{X}$ with $\lambda >0$; under $T$, the $\widehat{X}$-trajectories are the images under $F$ of the $-{\mathrm{\partial }}_s$-trajectories. Take $x=T(q,a)\in f^{-1}(a)$ and follow the $X^{\mathrm{\prime }}$-trajectory backwards, or equivalently the normalized ascending field $-X^{\mathrm{\prime }}\mathrm{/}\lambda$: in $(q,s)$-coordinates it runs through $F^{-1}(q,a)=(q,a)$ because $F$ is the identity near $s=a$, then through $(q,a+u)$ at ascending normalized parameter $u$, then through $F(q,a+u)=(h_{\alpha (a+u)}(q),a+u)$. At ascending normalized time $u=b-a$ this is $(h_1(q),b)=h_1(q)\in f^{-1}(b)$. Since $\phi (x)=q$, the transport of $X^{\mathrm{\prime }}$ from level $a$ to level $b$ is exactly $h_1\circ \phi$.