I'm trying to compute the derivative of the map $f:\Sigma\times [0,\delta)\to M$ given by $$f(p,t)=\exp_p tN(p),$$ in $X\in T_p\Sigma$, where $(M^n,g)$ is a Riemannian manifold, $\Sigma\subset M$ a immersed hypersurface and $N$ is a unit normal vector field along to $\Sigma$. If $X\in T_p \Sigma$ then I think something like that
\begin{align} df_{(p,t)}X &=\frac{d}{ds}\bigg|_{s=0}f\circ c(s) \end{align}
where $c$ is a differentiable curve in $\Sigma\times [0,\delta)$ with $c(0)=(p,t)$ and $c'(0)=X$. Taking $c(s)=(\exp_p sX,t)=(c_1(s),t)$, we have
\begin{align} df_{(p,t)}X &=\frac{d}{ds}\bigg|_{s=0}\exp_{c_1(s)}tN(c_1(s))\\ &\overset{*}{=}d(\exp_{c_1(s)})_{tN(c_1(s))}\nabla_{c'(s)}N(c(s))\bigg|_{s=0}\\ &= d(\exp_p)_{tN(p)}\nabla_X N(p). \end{align}
In fact, I would like know if the equality (*) make sense because in this calculation when I derive $\exp_{c_1(s)}tN(c_1(s))$ the curve $c_1$ is varying.
NOTE: The original problem is decompose the pullback metric. I posted here but I don't understand the solution suggested. So, I tried to do something different.