Let $\partial M$ denote the boundary of a $k$-dimensional embedded $C^1$-submanifold $M$ of $\mathbb R^d$, $T_x(\partial M)$ and $N_x(\partial M)$ denote the tangent and normal field of $\partial M$ at $x\in\partial M$, respectively, $\nu:\partial M\to\mathbb R^k$ denote the normal field and $\operatorname P_{T_x(\partial M)}$ denote the orthogonal projection of $\mathbb R^k$ onto $T_x(\partial M)$ for $x\in\partial M$.
Now let \begin{align}\frac{\partial f}{\partial\nu}(x)&:={\rm D}f(x)\nu(x)=\langle\nabla f(x),\nu(x),\\\nabla_{\partial M}f(x)&:=\operatorname P_{T_x(\partial M)}\nabla f(x),\\{\rm D}_{\partial M}f(x)&:={\rm D}f(x)\circ\operatorname P_{T_x(\partial M)},\\\nabla_{\partial M}\cdot f&:=\operatorname{tr}{\rm D}_{\partial M}f(x)\end{align} for $f\in C^1$ and $\kappa:=\nabla_{\partial M}\cdot\nu$.
How can we show that, if $f\in C^2(\mathbb R^k,\mathbb R)$, $$\Delta_{\partial M}:=\nabla_{\partial M}\cdot\nabla_{\partial M}f=\Delta f-\kappa\frac{\partial f}{\partial\nu}-\frac{\partial^2f}{\partial\nu^2}?\tag1$$
I was able to show that $$\frac{\partial^2f}{\partial\nu^2}=({\rm D}^2f(x)\nu(x))\nu(x)+{\rm D}f(x){\rm D}\nu(x)\nu(x)\tag2$$ and \begin{equation}\begin{split}&{\rm D}\left(\frac{\partial f}{\partial\nu}\nu\right)(x)y=(({\rm D}^2f(x)y)\nu(x))\nu(x)\\&\;\;\;\;\;\;\;\;\;\;\;\;+({\rm D}f(x){\rm D}\nu(x)y)\nu(x)+\frac{\partial f}{\partial\nu(x)}{\rm D}\nu(x)y\end{split}.\tag3\end{equation}
However, what I obtain from that is \begin{equation}\begin{split}&\Delta_{\partial M}f(x)=\Delta f(x)-\left(\nabla\cdot\frac{\partial f}{\partial\nu}\nu\right)(x)-\langle{\rm D}(\nabla f)\nu(x),\nu(x)\rangle\\&\;\;\;\;\;\;\;\;\;\;\;\;+\frac{\partial^2f}{\partial\nu^2}(x)+\frac{\partial f}{\partial\nu}(x)\langle{\rm D}\nu(x)\nu(x),\nu(x)\rangle\end{split},\tag4\end{equation} which seems to be wrong since even the sign of $\frac{\partial^2f}{\partial\nu^2}(x)$ is different ...