Let $(\mathcal{M},g)$ be a four-dimensional Lorentzian manifold and $\Sigma$ a Riemannian hypersurface such that $\mathcal{M}=\mathbb{R}\times\Sigma$. How to show (if true) that
$$T^{\ast}\mathcal{M}\vert_{\Sigma}\cong (\Sigma\times\mathbb{R})\oplus T^{\ast}\Sigma,$$
where $\Sigma\times\mathbb{R}$ denotes the trivial bundle. My motivation for this comes as follows: Lets take a section of $T^{\ast}\mathcal{M}$, which is a $1$-form $A=A_{\mu}\mathrm{d}x^{\mu}$. Then, I can naturally write $A=A_{0}\mathrm{d}t+A_{\Sigma}$ (where $t$ is the "time"-coordinate corresponding to $\mathbb{R}$), where $A_{0}\in C^{\infty}(\mathcal{M})$ and $A_{\Sigma}$ is a "time-dependent" $1$-form on $\Sigma$, i.e. a one-parameter family in $\Omega^{1}(\Sigma)$ labelled by the time-parameter $t$. Now, if I restrict this section to $\Sigma$, then I naturally get a map $$\Gamma^{\infty}(\Sigma,T^{\ast}\mathcal{M})=\Gamma^{\infty}(T^{\ast}\mathcal{M}\vert_{\Sigma})\xrightarrow{\cong}C^{\infty}(\Sigma)\oplus \Gamma^{\infty}(T^{\ast}\Sigma)\\ A=A_{0}\mathrm{d}t+A_{\Sigma}\mapsto (A_{0},A_{\Sigma}),$$
so I would expect the above splitting on the level of boundles.
For smooth manifolds $X$ and $Y$, we have $T^*(X\times Y) \cong \pi_1^*T^*X\oplus\pi_2^*T^*Y$ where $\pi_1 : X\times Y \to X$ and $\pi_2 : X\times Y \to Y$ are the natural projection maps. Now fix $x_0 \in X$ and consider the map $i : Y \to X\times Y$ given by $i(y) = (x_0, y)$. Note that $\pi_1\circ i = c_{x_0}$, the constant map $Y \to X$, $y \mapsto x_0$, and $\pi_2\circ i = \operatorname{id}_Y$. Therefore
\begin{align*} T^*(X\times Y)|_{i(Y)} &\cong i^*T^*(X\times Y)\\ &\cong i^*(\pi_1^*TX\oplus\pi_2^*TY)\\ &\cong i^*\pi_1^*TX\oplus i^*\pi_2^*TY\\ &\cong (\pi_1\circ i)^*TX\oplus(\pi_2\circ i)^*TY\\ &\cong c_{x_0}^*TX\oplus\operatorname{id}_Y^*TY\\ &\cong (Y\times\mathbb{R}^{\dim X})\oplus TY. \end{align*}
In your case $\mathcal{M} = \mathbb{R}\times \Sigma$, so $T^*\mathcal{M}|_{i(\Sigma)} \cong (\Sigma\times\mathbb{R})\oplus T\Sigma$. In fact, because $T\mathbb{R}$ is trivial, you have $T^*\mathcal{M} \cong (\mathcal{M}\times\mathbb{R})\oplus\pi_2^*T\Sigma$.