Let $M$ be a submanifold of $\mathbb R^l$ with the induced metric. Let $(\xi_\alpha)$ be the standard orthonormal basis on $\mathbb R^l$. For each $x \in M$, let $P_\alpha(x)$ the projection of $\xi_\alpha$ to $T_xM$.
Then we have shown that $$\Delta_M = \sum_{\alpha=1}^l P_\alpha^2,$$ where $\Delta_M$ denotes the Laplace-Beltrami operator.
The problem arises in the next Corollary to the Theorem.
With the same notation, we have $$\Delta_M f = \sum_{\alpha=1}^l \nabla^2 f(P_\alpha, P_\alpha).$$
Proof. Let $(X_i)$ be an orthonormal basis for $T_xM$. From $X_i = \langle X_i, P_\alpha\rangle P_\alpha$ and $\Delta_M f = \mathrm{tr}\nabla^2 f = \sum_{i\leq d}\nabla^2 f(X_i,X_i)$, $$\Delta_M f = \sum_{i\leq d}\nabla^2 f(X_i,X_i) = \sum_{i\leq d}\sum_{\alpha,\beta \leq l} \nabla^2(P_\alpha, P_\beta) \langle X_i, P_\alpha \rangle\langle X_i,P_\beta \rangle.$$ We have $$\sum_{\beta\leq l}\sum_{i\leq d} \langle X_i, P_\alpha\rangle \langle X_i, P_\beta\rangle P_\beta \overset{(*)}= \sum_{\beta \leq l} P_\beta \langle P_\alpha, P_\beta\rangle = P_\alpha.$$
Why do we get the equivalence...
(1) $X_i = \langle X_i, P_\alpha\rangle P_\alpha$. I think it is some kind of Gram-Schmidt argument?
(2) $(*)$?
As for $(1)$, since $\xi_1,\ldots,\xi_l$ is an orthonormal basis of $\mathbb{R}^l$, we have$$X_i=\sum_{\alpha=1}^l\langle X_i,\xi_\alpha\rangle \xi_\alpha$$for $i=1,\ldots,d$. Projecting both sides to $T_xM$, we get$$X_i=\sum_{\alpha=1}^l\langle X_i,\xi_\alpha\rangle P_\alpha=\sum_{\alpha=1}^l\langle X_i,P_\alpha\rangle P_\alpha.$$
We now turn to $(2)$. Since $X_1,\ldots,X_d$ is an orthonormal basis of $T_xM$, we have $$\langle u,v\rangle=\sum_{i=1}^d\langle X_i,u\rangle\langle X_i,v\rangle$$for every $u,v\in T_xM$. The equality $(*)$ is obtained by taking $u=P_\alpha$ and $v=P_\beta$.