Why are the restrictions uniformly Lipschitz continuous?

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First, some definitions.

We let $\overline{\mathbb{R}}=[-\infty,+\infty]$ be equipped with the differentiable structure which makes $\phi:\overline{\mathbb{R}} \to [-1,1]$ given by $\phi(t)=\frac{t}{\sqrt{1+t²}}$ be a diffeomorphism. A proposition in Matthias Schwarz's - Morse Homology states, in particular, the following.

Given $\xi,\eta $ smooth vector bundles over $\overline{\mathbb{R}}$ and $f: \xi \to \eta$ a smooth bundle map such that $f(0_{\pm \infty})=0_{\pm \infty}$, then its fiber restrictions $f_t$ are in particular Lipschitz continuous, uniformly in $t \in \mathbb{R}$.

I don't understand why that must be true. Analyzing in the chart, it seems pretty easy to come up with a smooth "bundle" function $f: [-1,1] \times \mathbb{R}^n \to [-1,1] \times \mathbb{R}^m$ satisfying the hypothesis which is not Lipschitz continuous uniformly in $t$: take a non-Lipschitz smooth function $g: \mathbb{R}^n \to \mathbb{R}^m$, a bump function $\lambda: [-1,1] \to \mathbb{R}$ supported in $[-1/2,1/2]$ and consider $f(t,x)=(t,\lambda(t)g(x)).$

What am I missing here? Thanks in advance.