In Abraham and Marsden's Foundations of Mechanics, they prove Morse lemma via Moser's trick. They are able to reduce the proof so that it suffices to find a smooth family of vector fields $Z_t$ such that $$\iota_{Z_t}\omega_t+(f-g)=0, \quad Z_t(0)=0,$$ where $f$ is the function under consideration in Morse lemma statement, $g(x)=\frac{1}{2}D^2f(0)(x,x)$ and $\omega_t=tdf+(1-t)dg$.
Now, the book says that it is "easy to see that $Z_t$ exists near $0$ by the nondegeneracy hypothesis", which I assume is referring to the non-degeneracy of the hessian $D^2f(0)$.
However, I can't see how it is easy to see that. For instance, $\omega_t(0)$ is $0$, so $Z_t(0)$ could be whatever we want and I see no way to easily reconcile this*. Furthermore, $\omega_t$ is a $1$-form and we want a $Z_t$ such that $\omega_t(Z_t)$ is some number. This gives a lot of redundancy for $Z_t$.
My question is: How to build such $Z_t$?
*This is quite different from the usage of Moser's trick on the symplectic situations (Darboux's theorem, say), where $\omega_t$, when constant on a region of interest (like the origin in this case), is usually non-degenerate (not to mention being a $2$-form, which is relevant for the next thing I am about to say)
For those not familiar, Morse lemma is as follows.
Let $f:M \to \mathbb{R}$ be such that $\mathrm{Hess}_p$ is non-degenerate and $f(p)=0$. Then there exists a chart $\phi$ around $p$ such that in local coordinates $$f(x)=D^2f(0)(x,x).$$
Since $df(0) = 0$, given a vector $v$ we can write $$ df(x)(v) = \big<G(x)v,x\big>, $$ where $$ G(x) = \int_0^1 \mathrm{Hess}\;f(sx)\;ds. $$ In fact, $$ df(x)(v) = \int_0^1 \frac{d}{ds}df(sx)(v)\;ds = \int_0^1 D^2f(sx)(v,x)\;ds = \big<\big(\int_0^1\mathrm{Hess}\;f(sx)\;ds\big)v,x\big>. $$ Thus, the equation $$ i_{Z_t}\omega_t = g - f $$ becomes $$ \big<(tG(x) + (1 - t)H)Z_t(x),x\big> = \big<((1/2)H - G(x))x,x\big> $$ where $H = \mathrm{Hess}\;f(0)$. Whence, it suffices to find $Z_t$ such that $$ (tG(x) + (1 - t)H)Z_t(x) = ((1/2)H - G(x))x. $$ Since $G(0) = H$, the operator $(tG(x) + (1 - t)H)$ is invertible for $x$ near $0$. Define $$ Z_t(x) = (tG(x) + (1 - t)H)^{-1}((1/2)H - G(x))x $$ in a neighbourhood of $0$.