This has nothing to do with minimality of $\Sigma$.

By definition of the exponential map, the mapping $\gamma_p(t) = \phi(p,t)$, as a map $[0,\epsilon)\to M$ is a geodesic ray, with unit speed, initial position $p\in \Sigma$, and initial velocity $N_p$.

Let $V$ be a vector in $T_p M$ extended to $\Sigma \times [0,\epsilon)$ along $\gamma_p(t)$ by the product structure. Equivalently, you have that $V$ is Lie-transported along $\gamma_p(t)$, or that $[\dot{\gamma_p}, V] = 0$. (Vector field commutators are independent of metric!)

The pushforward of $V$ is a vector field that satisfies $V(t) = 0$. To get your splitting it suffices to show that $\langle V, \dot{\gamma_p}\rangle_g = 0$. But this follow because at time $t = 0$ the equality is true, and $$ \nabla_{\dot{\gamma_p}} \langle V, \dot{\gamma_p}\rangle_g = \langle \nabla_V \dot{\gamma_p}, \dot{\gamma_p}\rangle_g = 0 $$ by construction and geodesy.

This has nothing to do with minimality of $\Sigma$.

By definition of the exponential map, the mapping $\gamma_p(t) = \phi(p,t)$, as a map $[0,\epsilon)\to M$ is a geodesic ray, with unit speed, initial position $p\in \Sigma$, and initial velocity $N_p$.

Let $V$ be a vector in $T_p \Sigma$ extended to $\Sigma \times [0,\epsilon)$ along $\gamma_p(t)$ by the product structure. Equivalently, you have that $V$ is Lie-transported along $\gamma_p(t)$, or that $[\dot{\gamma_p}, V] = 0$. (Vector field commutators are independent of metric!)

The pushforward of $V$ is a vector field that satisfies $V(t) = 0$. To get your splitting it suffices to show that $\langle V, \dot{\gamma_p}\rangle_g = 0$. But this follow because at time $t = 0$ the equality is true, and $$ \nabla_{\dot{\gamma_p}} \langle V, \dot{\gamma_p}\rangle_g = \langle \nabla_V \dot{\gamma_p}, \dot{\gamma_p}\rangle_g = 0 $$ by construction and geodesy.