To show that, $\nabla_{\mathbb{H}^n}^2x_i=x_ig_{\mathbb{H}^n}, 1\leq i\leq n$ and $\nabla_{\mathbb{H}^n}^2t=tg_{\mathbb{H}^n}$. 
We regard the hypobilic space $\mathbb{H}^n$ as a submanifold of the standard Minkowski space, i.e.,
$$
\mathbb{H}^n:=\{(x,t):|x|^2-t^2=-1\}.
$$
To show that,
$$
\nabla_{\mathbb{H}^n}^2x_i=x_ig_{\mathbb{H}^n}, 1\leq i\leq n
$$
and
$$
\nabla_{\mathbb{H}^n}^2t=tg_{\mathbb{H}^n}.
$$

The positive-curvature manifold, standard sphere shares the similar property. By calculating some Lie derivatives, we can obtain that for $S^n=\{(x_1, \cdots, x_n)\in\mathbb{R^{n+1}}:|x|^2=1\}$,
$$
\nabla_{S^n}x_i+x_ig_{S^n}=0, 1\leq i\leq n+1.
$$
But I wonder how to prove for the hyperbolic space. Appreciate any help!.
 A: $\DeclareMathOperator{\llangle}{\langle\!\langle} \DeclareMathOperator{\rrangle}{\rangle\!\rangle}
\newcommand{\H}{\Bbb H^n}
\DeclareMathOperator{\hess}{Hess}
\DeclareMathOperator{\grad}{grad}$
I am sure there is a coordinate-dependent way that might be shorter than the proof I will give, but I really like the following coordinate-free proof.
The exact same proof also works in the case of the sphere, with obvious changes of the ambient inner product and of some signs.
Let $\Bbb R^{n,1} = (\Bbb R^{n+1},\llangle \cdot,\cdot\rrangle)$, where $\llangle (x,t),(y,s) \rrangle = \langle x,y\rangle - ts$, with $x,y \in \Bbb R^n$, $t,s\in \Bbb R$ and $\langle\cdot ,\cdot \rangle$ the usual inner product on $\Bbb R^n$.
Let $Q$ be the associated quadratic form, so that $\H = \left(Q^{-1}(-1)\cap \{t>0\}, g \right)$, where $g$ is the restriction of $\llangle\cdot,\cdot\rrangle$ to the tangent spaces.
Let $\nabla$ be the ambient flat Levi-Civita connection, which is given by the usual directional derivative, and let $\nabla^{\H}$ be that of $\H$.
For $p\in \H$, we have the orthogonal splitting
$$
\Bbb R^{n,1} = \Bbb R p \oplus T_p\H,
$$
in which any vector $v$ reads as
$$
v = -\llangle v,p\rrangle p + \left( v + \llangle v,p\rrangle p\right).
$$
It follows that the Levi-Civita connection for $\H$ is given by
$$
\nabla^{\H}_XY(p) = \nabla_XY(p) + \llangle \nabla_XY(p),p\rrangle p.
$$
Let $\{e_1,\ldots,e_{n+1}\}$ be the canonical basis of $\Bbb R^{n,1}$.
The coordinate functions $\{x^1,\ldots,x^n,t\}$ on $\Bbb R^{n,1}$ are given by
\begin{align}
x^i(p)& = \llangle p,e_i\rrangle,\quad i\in\{1,\ldots,n\}, \\\\
t(p)& = -\llangle p,e_{n+1}\rrangle,
\end{align}
and for $Y$ a tangent vector, their differentials are simply given by
\begin{align}
dx^i(Y) &= \llangle Y,e_i\rrangle,\quad i\in\{1,\ldots,n\}, \\\\
dt(Y) &= -\llangle Y,e_{n+1}\rrangle.
\end{align}
If $f$ is a function on $\H$, then by definition,
$$
\hess f = \nabla^{\H} df,
$$
that is, for $X$ and $Y$ vector fields on $\H$,
$$
\hess f(X,Y) = X(df(Y)) - df(\nabla^{\H}_XY).
$$
Hence, for $i\in \{1,\ldots,n\}$, $p\in \H$ and $X$, $Y$ vector fields on $\H$, we have
\begin{align}
\hess x^i(X,Y) &= X\left(dx^i(Y)\right) - dx^i(\nabla^{\H}_XY)\\
&= X \llangle Y,e_i\rrangle - \llangle \nabla^{\H}_XY,e_i\rrangle\\
&= X \llangle Y,e_i\rrangle - \llangle\nabla_XY + \llangle \nabla_XY,p\rrangle p , e_i \rrangle\\
&= \underbrace{X\llangle Y,e_i\rrangle - \llangle \nabla_XY,e_i\rrangle}_{=\llangle Y,\nabla_X e_i\rrangle =0} - \llangle \nabla_XY,p\rrangle \underbrace{\llangle p,e_i\rrangle}_{=x^i}\\
&= -x^i \llangle \nabla_XY,p\rrangle.
\end{align}
Similar calculations, taking care of the fact that $t = -\llangle\cdot, e_{n+1}\rrangle$, also show that
$$
\hess t(X,Y) = -t \llangle \nabla_XY,p\rrangle.
$$
It now all boils down to show that $\llangle \nabla_XY(p),p\rrangle = -g(X,Y)$ for $X$ and $Y$ vector fields on $\H$.
Since $\llangle Y,p\rrangle = 0$ (recall that for all $p$, $Y(p)\in\{p\}^{\perp}$), then
$$
\llangle \nabla_XY,p\rrangle = X\underbrace{\llangle Y,p\rrangle}_{=0} - \llangle Y,\nabla_Xp\rrangle,
$$
when $p$ is thought of as the position vector field.
Since $\nabla$ is the usual directional derivative, it follows that $\nabla_Xp = X$.
Hence
$$
\llangle \nabla_XY,p\rrangle = -\llangle Y,X\rrangle = -g(Y,X) = -g(X,Y).
$$
Remark: this implies that the ambient coordinates $\{x^1,\ldots,x^n,t\}$ are eigenfunctions of the Laplacian $\Delta_{\H}$, with eigenvalue $n$.
More generally, so is any function $f(p) = \llangle p,a\rrangle$ with $a\in \Bbb R^{n,1}$ a fixed nonzero vector.

Alternatively, you can notice that $\grad(x^i) = e_i$ and $\grad(t) = -e_{n+1}$, which yield
\begin{align}
\grad^{\H}(x^i) &= \grad(x^i) + \llangle \grad(x^i),p \rrangle p\\
&= e_i + x^i p,\\\\
\grad^{\H}(t) &= \grad(t) + \llangle \grad(t),p\rrangle p \\
&= -e_{n+1} +tp.
\end{align}
Since for any function $f$, we have
$$
\frac{1}{2}\mathcal{L}_{\grad^{\H}(f)}g= g(\nabla^{\H}\grad^{\H}(f)\cdot,\cdot) = \hess f,
$$
it remains to compute $\mathcal{L}_{\grad^{\H}(x^i)}g$ and $\mathcal{L}_{\grad^{\H}(t)}g$ thanks to the latter expression.
The computations will be substantially identical than that of the first proof.
