Let $d\in\mathbb N$, $k\in\{1,\ldots,d\}$ and $M$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary.
Can we prove the usual identity (in the case, where $M$ is an open subset of $\mathbb R^d$) for the divergence of the product of a scalar field and a vector field on $M$?
To be precise, if $N\supseteq M$ is an embedded $C^1$-submanifold with boundary, $\operatorname P_N(a)$ is the orthogonal projection of $\mathbb R^d$ onto $T_a\:N\supseteq T_a\:M$ for $a\in N$, $E$ is a $\mathbb R$-Banach space and $f:N\to E$ is $C^1$-differentiable at $a\in M$, define $${\rm D}_Mf(a):=T_a(f)\circ\operatorname P_M(a)\tag4.$$ If $E=\mathbb R$, define $$\langle\nabla f,v\rangle:=T_a(f)v\;\;\;\text{for }v\in T_a\:N\tag5$$ and $^1$ $$\nabla_Mf(a):=\operatorname P_M\nabla f(a)\tag6.$$ And if $E=\mathbb R^d$, define $$(\operatorname{div}f)(a)=\operatorname{tr}T_a(f)\;\;\;\text{for all }a\in M\tag7$$ and $$(\operatorname{div}_Mf)(a):=\operatorname{tr}{\rm D}_Mf(a)\tag8.$$
Now let $f:M\to\mathbb R$ and $X:M\to\mathbb R^d$ both be $C^1$-differentiable at $a\in M$ and $(\tau_1,\ldots,\tau_k)$ be an orthonormal basis of $T_a\:M$. Then we should have $$(\operatorname{div}fX)(a)=\sum_{i=1}^k\langle T_a(fX)\tau_i,\tau_i\rangle=\langle\nabla f(a),\operatorname P_M(a)X(a)\rangle+f(a)(\operatorname{div}X)(a)\tag9,$$ where $$\langle\nabla f(a),\operatorname P_M(a)X(a)\rangle=\langle\nabla f(a),X(a)\rangle\tag{10}.$$
So, the usual formula $$(\operatorname{div}fX)(a)=\langle\nabla f(a),X(a)\rangle+f(a)(\operatorname{div}X)(a)\tag{11}$$ should actually hold.
What's worrying me is that I've read that we would need to assume $X(a)\in T_a\:M$ for that, but I don't see why we should need this assumption since $X(a)$ should automatically be projected onto $T_a\:M$ by $\langle\nabla f(a),X(a)\rangle$ anyways.
$^1$ Note that $${\rm D}_Nf(a)v=\langle\nabla_Nf(a),v\rangle\;\;\;\text{for all }v\in\mathbb R^d\tag{6'}.$$ Moreover, if $M=N$, then $\nabla f_N(a)=\nabla f(a)$.
$^2$ Note that if $M=N$, then $(\operatorname{div}_Mf)(a)=(\operatorname{div}f)(a)$.