\textbf{Geometric Measure Theory}\\
Derivatives of functions and vector fields on hypersufaces can also be defined in terms of projections from $\mathbb{R}^{n+1}$ onto the tangent space of $M$. This is the framework used in geometric measure theory where coordinate systems are not available. The notions defined below can be adapted to countable $n-$ rectifiable subsets of $\mathbb{R}^{n+1}$.
\[\nabla^Mf(x)=Df(x)-(Df,\nu(x))\nu(x)\]
for $x\in M$ where $Df$ denotes the usual gradient of $f$ in $\mathbb{R}^{n+1}$ (or of its extension into $\mathbb{R}^{n+1}$).
This can also be written as
\[\nabla^Mf(x)=\sum_{i=1}^n\tau_iD_{\tau_i}f(x)\]
where $\tau_1,...,\tau_n$ form a orthogonal basis of $T_xM$.
The Laplace-Beltrami operator on $M$ of a twice differentiable function $f$ is defined by
\begin{align*}
\Delta_Mf&=div_M(grad^M\ f),
\end{align*}
\begin{align*}
\vec{H}&=-H\nu=-div_M(\nu)\nu,
\end{align*}
\begin{align*}
\Delta_Mf&=\Delta_{\mathbb{R}^{n+1}}f-D^2f(\overrightarrow{\nu},\overrightarrow{\nu})-div_M(\overrightarrow{\nu})\overrightarrow{\nu}\cdot Df\\
&=\Delta_{\mathbb{R}^{n+1}}f-D^2f(\vec{\nu},\vec{\nu})+\overrightarrow{H}\cdot Df
\end{align*}
It follows that
\[\Delta_M x_i=\overrightarrow{H}\cdot \mathbf{e_i},\]
\[\Delta_M\mathbf{X}=\overrightarrow{H}.\]
It is also easy to check that ($X, \mathbf{e_i}$ are a column vectors)
\begin{align}
\delta_if&=D_if-(Df,\nu)\nu_i,\ i=1,...,n+1,\\
\nabla^Mf=\delta f&=(I_{n+1}-\nu\otimes\nu)Df,\ \mathbf{Orthogonal\ Projection},\ \\
D_{\tau}X&=DX\tau=D_jX^i\tau_j\mathbf{e_i},\ \tau=\sum^n_{j=1}\tau_j\mathbf{e_j}\in T_pM,\ \\
div_M(X)&=\sum_{i=1}^n\delta_i(X^i)=\sum_{i=1}^n\tau_i\cdot D_{\tau_i}X=Trace((I_{n+1}-\nu\otimes\nu)DX)\nonumber\\
&=div_{\mathbb{R}^{n+1}}DX-\nu^T DX\nu,\ \ where \ X=\sum_{i=1}^{n+1}X^i\mathbf{e}_i\\
H&=\sum_{i=1}^n\delta_i(\nu^i),\ where\ \ \nu=\sum_{i=1}^{n+1}\nu^i\mathbf{e}_i,\\
\Delta_Mf&=\sum_{i=1}^n\delta_i(\delta_if),\nonumber\\
&=div_{M}(Df)+\overrightarrow{H}\cdot Df,\nonumber\\
&=\Delta_{\mathbb{R}^{n+1}}f-D^2f(\vec{\nu},\vec{\nu})+\overrightarrow{H}\cdot Df
\end{align}
It is also clear that
\begin{align*}
\Delta_{M} f&=div_{M}(Df-<\nu,Df>\nu)\\
&=div_{M}(Df)-<\nu,Df>\delta_i\nu^i\\
&=div_{M}(Df)+\vec{H}\cdot Df.
\end{align*}
Naturally, we can imagine that
\begin{equation*}
\nabla^M_{\tau}X=DX\tau-<DX\tau,\nu>\nu,
\end{equation*}
For any $\xi,\eta\in T_pM$, let
\begin{align*}
B(\xi,\eta)&=-\langle D_{\xi}\nu,\eta\rangle\\
&=\xi^iD_{e_i}\nu^j\eta_j\\
&=\xi^i(\delta_i\nu^j+\nu^i\nu^hD_hv^j)\eta_j\\
&=\xi^i\delta_i\nu^j\eta_j
\end{align*}
By using the sign distance function with $\nu=Dd$ (See E.Giusti, Chapter 8) or level set method with $\nu=\frac{DF}{|DF|}$,
it is clear that
\begin{equation}
D_i\nu^j=D_j\nu^i,\ \ \ \ \ \delta_i\nu^j=\delta_j\nu^i,
\end{equation}
then $B(\xi,\eta)$ is symmetric with respect to $\xi,\eta\in T_pM$.
By the GMT definition, we can also define the corresponding operators for codimension bigger than one case.
In the mean curvature flow case, if $h(p,t)=f(x,t)$ for $x=F(p,t)$. The partial derivative of $h$ with respect to time, equals the total time derivative of $f$ along ($M_t$), that is,
\begin{equation}
\frac{\partial h}{\partial t}(p,t)=\frac{d f}{d t}(x,t)=\frac{d f}{d t}(F(p,t),t)=\frac{\partial f}{\partial t}(x,t)+Df(x,t)\cdot \frac{\partial F}{\partial t}(p,t)
\end{equation}
for $x=F(p,t)$ or in short
\begin{equation}
\frac{\partial h}{\partial t}=\frac{d f}{d t}=\frac{\partial f}{\partial t}+Df \cdot \vec{H}.
\end{equation}
Notice that
\begin{equation}
\Delta_{M_t}f
=\Delta_{\mathbb{R}^{n+1}}f-D^2f(\vec{\nu},\vec{\nu})+\overrightarrow{H}\cdot Df
\end{equation}
It follows that
\begin{align}
(\frac{d }{dt}-\Delta_{M_t})f&=\frac{\partial f }{\partial t}-div_{M_t}Df\nonumber\\
&=(\frac{\partial}{\partial t}-\Delta_{\mathbb{R}^{n+1}})f+D^2f(\vec{\nu},\vec{\nu})
\end{align}
for the difference between the ambient heat operator and the heat operator on $M_t$.
Recall that
\begin{align*}
(\frac{\partial}{\partial t}-\Delta_{M_t})h&=(\frac{d }{dt}-\Delta_{M_t})f.
\end{align*}
As a example, consider
\[f(x,t)=|x-x_0|^2+2n(t-t_0).\]
Since
\[\frac{\partial f}{\partial t}(x,t)=2n,\]
\[Df(x,t)=2(x-x_0)\]
and
\[div_{M_t}(x-x_0)=n,\]
we obtain
\begin{align*}
(\frac{d }{dt}-\Delta_{M_t})(|x-x_0|^2+2n(t-t_0))=0.
\end{align*}
As for the Monotonicity formula for functions on minimal surface $M$, we need the following simple fact:
\[\Delta_{M}(fg)=\Delta_{M}(f)g+f\Delta_{M}(g)+2\langle\nabla_M f,\nabla_M g \rangle\]
\[\Delta_{M} X=0,\]
\[\Delta_{M} |X|^2=2\langle\nabla_M X,\nabla_M X \rangle=2n.\]
\textbf{Second fundamental form for level set}\\
Let $\nu=\frac{DF}{|DF|}$, note that $grad F=DF$ in $R^{n+1}$. It follows that
\begin{align*}
II_{\nu}(X,Y)&=<\nu,D_XY>=\frac{1}{|DF|}<grad F,D_XY>\\
&=\frac{1}{|DF|}\nabla_XYF\\
&=\frac{1}{|DF|}\{-YXF+\nabla_XYF\}\\
&=-\frac{1}{|DF|}D^2F(X,Y),
\end{align*}
Hence,
\[II_\nu=-\frac{D^2F}{|DF|}.\]
\section{Gradient, Divergence, Laplace , Ricci Identity}
Let $G=det(g_{ij})$. We have
\[\frac{1}{\sqrt{G}}\cdot\frac{\partial \sqrt{G}}{\partial x_k}=\Gamma_{ki}^i.\]
By the definition of Christoffol symbols, we have
\[\Gamma_{ki}^i=\frac{1}{2}g^{il}\{\frac{\partial g_{il}}{\partial x_k}+\frac{\partial g_{kl}}{\partial x_i}-\frac{\partial g_{ik}}{\partial x_l}\}=\frac{1}{2}g^{il}\frac{\partial g_{il}}{\partial x_k}.\]
By using the differentiate w.r.t. determinant, we have
\[\frac{1}{\sqrt{G}}\cdot\frac{\partial \sqrt{G}}{\partial x_k}=\frac{1}{2G}\cdot\frac{\partial G}{\partial x_k}=\frac{g^{*}_{ij}}{2G}\cdot\frac{\partial g_{ij}}{\partial x_k}=\frac{g^{ij}}{2}\cdot\frac{\partial g_{ij}}{\partial x_k}.\]
This finish the proof.
Let $X\in\chi(M)$, we define
\[div(X):=C^1_1(DX).\]
In local coordinate, we have $X=X^i\frac{\partial}{\partial x_i}$. Then
\[DX=X^i_{,j}\frac{\partial}{\partial x_i}\otimes dx^j,\]
and
\[div(X)=\sum\limits_{i=1}^n X^i_{,i},\]
Notice that
\begin{align*}
&\quad DX(dx^i,\partial_j)\\
&=D_{\partial_j}{X}(dx^i)=\partial_j(X(dx^i))-X(D_{\partial_j}dx^i)\\
&=\frac{\partial X_i}{\partial x_j}+\Gamma_{jk}^iX^k
\end{align*}
Therefore, we have
\[div(X)=\frac{\partial X_i}{\partial x_i}+\Gamma_{ik}^iX^k\]
By using the previous argument, we have
\[div(X)=\frac{\partial X_i}{\partial x_i}+\frac{1}{\sqrt{G}}\frac{\partial\sqrt{G}}{\partial x_k}X^k=\frac{1}{\sqrt{G}}\frac{\partial}{\partial x_i}(\sqrt{G}X^i).\]
With the help of Riemmanian connection $g$ and $df$, we can define $grad f$, i.e., for any $X\in\chi(M)$,
\[g(grad f, X)=df(X)=X(f).\]
Obviously, $grad f$ is a tangent vector field of $M$.
In local coordinate, we have $df=\frac{\partial f}{\partial x_i}dx^i$. If $grad f= a^i\frac{\partial}{\partial x_i}$, we have
\[a^i\langle \frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\rangle=df(\frac{\partial}{\partial x_j})=\frac{\partial}{\partial x_j}(f).\]
Then we have
\[a^i g_{ij}=\frac{\partial f}{\partial x_j}.\]
It follows that
\[a^i =\frac{\partial f}{\partial x_j} g^{ji}.\]
Thus,
\[grad~f=\frac{\partial f}{\partial x_j}g^{ij}\frac{\partial}{\partial x_i}.\]
Now we can define
\[\Delta f:=div(grad~f).\]
In local coordinate, we have
\[\Delta f=div(grad~f)=\frac{1}{\sqrt{G}}\frac{\partial}{\partial x_i}(\sqrt{G}g^{ij}\frac{\partial f}{\partial x_j}).\]
We can also define $Hess(f)$, i.e.,
\[Hess(f)=D(df).\]
In local coordinate, we have
\begin{align*}
Hess(f)(X,Y)&=D(df)(X,Y)\\
&=D_Y(df)(X)=Y(df(X))-(df)(D_YX)\\
&=Y(X(f))-(D_YX)f\\
&=X(Y(f))-(D_XY)f\\
&=Hess(f)(Y,X).
\end{align*}
Hessian of f is a symmetric (0,2) type tensor.
\begin{definition}
Let $S$ be a (0,2) symmetric covariant tensor. We can define the trace of $S$ as
\[TrS=\sum\limits_{i=1}^nS(e_i,e_i),\]
where ${e_i}$ are standard orthogonal basis.
\end{definition}
Let $e_i=b_i^j\frac{\partial}{\partial x_j}$. Then
\[\delta_{kl}=\langle e_k,e_l\rangle=b_k^ib_l^j\langle \frac{\partial}{\partial x_k},\frac{\partial}{\partial x_l}\rangle=b_k^ib_l^jg_{ij}.\]
Let $C=(b^j_i)$. Then we have $CgC^T=I_n$. It is easy to see that $C^T$ is the inverse matrix of $Cg$. Hence $C^TCg=I_n$ and $g^{-1}=C^TC$. This is exactly the identity \[g^{ij}=b_k^ib^j_k.\]
This leads to
\[TrS=S(e_k,e_k)=S(b_{k}^i\frac{\partial}{\partial x_i},b_{k}^j\frac{\partial}{\partial x_j})=b_{k}^ib_{k}^jS(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j})=g^{ij}S(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}).\]
Recall P119 in Mei, we can also define\[dix(x):=\langle D_{e_i}X, e_i\rangle,\]
where ${e_i}$ is a normal orthogonal basis.
In the following, we can check that the two definitions of $\Delta$ in previous paragraph are coincident.
\proof~ Firstly, we have
\begin{align*}
\Delta f=g^{ij}Hess(f)(\partial x_i,\partial x_j)\\
=g^{ij}\frac{\partial}{\partial x_j}(\frac{\partial f}{\partial x_i})-g^{ij}\Gamma_{ij}^k\frac{\partial f}{\partial x_k}
\end{align*}