Showing that $k^*(H_f)\lrcorner k^*\sigma=k^*(H_f\lrcorner\sigma)$for symplectomorphism $k$, contraction $\lrcorner$ and symplectic product $\sigma$
Definitions:
Let $f:\mathbb{R}^n\to \mathbb{R}^n$ be a smooth mapping and $Jf$ be the Jacobian of $f$. Then if $\nu$ is an $m$-form on $\mathbb{R}^n$, the pullback of $\nu$ by $f$ is defined as $(f^*\nu)(w) = \nu(Jf w_1,\dots,Jf w_m)$ for $w \in (\mathbb{R}^n)^m$.
Let $\alpha = (x, \xi), \beta = (y, \nu) \in \mathbb{R}^{2n}$. Then the symplectic product between $\alpha$ and $\beta$ is defined as $\sigma(\alpha, \beta) = \left<\xi, y\right> - \left<\nu, x\right>$.
Let $k:\mathbb{R}^{2n}\to \mathbb{R}^{2n}$. Then $k$ is said to be a symplectomorphism, if $k^*\sigma = \sigma$.
Let $\nu$ is a differential $m$-form and $X$ be a vector field. Then, the contraction of $\nu$ by $X$ is defined as the $(m-1)$-form $(X\lrcorner \nu)(w) = \nu(X, w), w \in (\mathbb{R}^n)^{m-1}$.
Let $f$ be a smooth function defined on $\mathbb{R}^{2n}$. Then the Hamiltonian vector field $H_f$ is a vector field such that $\forall w \in \mathbb{R}^{2n}:\sigma(w, H_f) = df(w)$ where $d$ is the differential 1-form.
Preamble: I am trying to understand an intermediary step in a proof for Jacobi's theorem, that is $H_f = k_*(H_{k^*}f)$, where one needs the equality $k^*(H_f)\lrcorner k^*\sigma = k^*(H_f\lrcorner \sigma)$. If I write out the LHS I get $(k^*(H_f)\lrcorner k^*\sigma)(w) = k^*\sigma(k^*(H_f), w) = \sigma(k^*(H_f), w) = -\sigma(w, H_f \circ Jk)$ ($\sigma$ is antisymmetric), where I am not sure how to either "move" the $Jk$ or argue why $H_f \circ Jk$ is necessarily a Hamiltonian vector field.
The RHS evaluates nicely to
$(k^*(H_f\lrcorner \sigma))(w) = (H_f\lrcorner \sigma)(Jk w) = \sigma(H_f, Jk w) = -\sigma(Jk w, H_f) = -df(Jk w)$.
Question: Is there a neat way to show the equality $k^*(H_f)\lrcorner k^*\sigma=k^*(H_f\lrcorner\sigma)$? If there is, what is it? My current approach radiates guess-and-checkness rather than intelligent observations.
