The reference for the following discussion is Lee's Introduction to Riemannian Manifolds. As can be seen in chapter 8 of this book, the mean curvature $H$ of a hypersurface $(M^{(n)},g)$ at $p\in M$ is defined as $H=\frac{1}{n}\mathsf{tr}(s)$, where $s$ is the shape operator (Weingarten map) of $M$ at $p$ in the direction of a smooth unit normal vector field $N$. Since $s$ is a self-adjoint operator on the tangent space $T_p M$, linear algebra assures us that there exists an orthonormal basis $\{E_1,\ldots,E_n\}$ for $T_p M$ consisting of eigenvectors of $s$. This makes it relatively easy to diagonalize $s$. If the eigenvalues corresponding to $E_1,\ldots,E_n$ are $\lambda_1,\ldots,\lambda_n$ respectively, then the representation of $s$ in $\{E_1,\ldots,E_n\}$ is simply $\mathsf{diag}(\lambda_1,\ldots,\lambda_n)$, thereby leading to $$H=\frac{1}{n}(\lambda_1+\ldots+\lambda_n).$$ Now I'd like to know how to start from this expression to get $$H=\frac{1}{n}(\mathsf{tr}_g h)(p)$$ if $\mathsf{tr}_g h$ is the trace of the scalar second fundamental form $h$ with respect to $g$. The relation between $h$ and $s$ is given by $$\left<s(X),Y\right>_g=h(X,Y)\tag{$*$}$$ for $X,Y\in\mathfrak{X}(M)$. In the above equation, it is essential to know that $s$ is a map from $\mathfrak{X}(M)$ to $\mathfrak{X}(M)$.
I'm not sure if I gave enough information to begin our discussion. Feel free to ask more if you don't have Lee's IRM. Thank you.
Update(2022/2/2):
Here is an attempt resulting from my understanding of $\mathsf{tr}_g h$. Following Lee's convention, we denote the dual basis of $\{E_1,\ldots,E_n\}$ by $\{\epsilon^1,\ldots,\epsilon^n\}$. Then we write $h_p=h_{ij}\epsilon^i\otimes\epsilon^j$ in order to compute $$(\mathsf{tr}_g h)(p)=g^{ik}(p)h_{ik}.$$ You can see p. 28 of IRM for more information about this formula. Now $$\begin{align} g^{ik}(p)h_{ik}&=g^{ik}(p)h_p(E_i,E_k)\\ &=g^{ik}(p)\left<s(E_i),E_k\right>_g\tag{bringing ($*$) in}\\ &=g^{ik}(p)\left<\lambda_i E_i,E_k\right>_g\\ &=\sum_{i=1}^n g^{ii}(p)\lambda_i. \end{align}$$ Look at the last equality. Those redundant $g^{ii}(p)$'s put me in a very awkward position! How can I get rid of them?
Update(2022/2/3):
A kind person reminded me that $g^{ij}(p)=\delta_{ij}$ due to the orthonormal basis. The job is done!
This is just some definitions plus linear algebra, since the involved quantities are either vector or tensor fields and their values can be evaluated pointwise. To an eigenvalue $\lambda$ of $S$ you have by definition a nonzero vector $v$ such that $Sv = \lambda v$, so that $$g(Sv, w) = \lambda g(v,w)$$ The trace of the second fundamental form wrt to $g$ is just (possibly up to a factor depending on the definition you are using, like $\frac{1}{n}$) $$\sum_i h(e_i, e_i) =\sum_i g(Se_i, e_i)$$ for any $g$ orthormormal basis $\{e_i:i=1,\dots,n\}$. Now plug in the basis of the eigenvectors wrt to $g$, which is orthonormal to see that this is just the sum of the eigenvalues. The existence of the latter is something you should know from linear algebra. The fact that the trace is welldefined and does not depend on the orthonormal basis is an exercise in linear algebra, too, which you should do on your own if this is not known to you.