Let $d\in\mathbb N$ and $M\subseteq\mathbb R^d$ be bounded and open such that $\partial M$ is of class $C^1$ (i.e. a $(d-1)$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$).
If $f:\partial M\to\mathbb R$ is $C^1$-differentiable, we can find the following definition of the "tangential gradient" of $f$ in Shapes and Geometries: Metrics, Analysis, Differential Calculus, and Optimization, Second Edition (p. 492)$^1$:
Why is it important to consider a $C^1$-extension $F$ of $f$ on a tubular neighborhood (or even on that specific one)? Why can't we take any $C^1$-extension of $f$, i.e. any $\tilde f\in C^1(O)$, where $O$ is an $\mathbb R^d$-open neighborhood of $\partial M$, with $$f=\left.\tilde f\right|_{\partial M}?\tag1$$ Is it needed to show that $g(F)$ is well-defined, i.e. independent of the choice of $F$? In any case, how can we show that it actually is well-defined?
EDIT: Meanwhile, I've found others references which consider arbitrary $C^1$-extensions. But it's still not clear to me why the definition of the tangential gradient is independent of the choice of the extension.
$^1$
Partial answer:
Consider the setting of my other question: Does the pushforward of a smooth map on a manifold coincide with the derivative in a tangent direction of any local extension?.
In that question, I've proven (hopefully without making any mistake) that if $v\in T_x\:M$ and $\gamma$ is a $C^1$-curve on $M$ through $x$ with $\gamma'(0)=v$, then $${\rm D}_vf(x):=(f\circ\gamma)'(0)\in T_{f(x)}\:E\tag2$$ is well-defined, i.e. independent of the choice of $\gamma$. Moreover, if $\tilde f$ is a $C^1$-extension of $f$ at $x$, then $${\rm D}_vf(x)={\rm D}_v\tilde f(x)={\rm D}\tilde f(x)v\tag3.$$
We can immediately infer the following: If $h\in\mathbb R^d$, then $$h=v+w\tag4$$ for some unique $(v,w)\in T_x\:M\times N_x\:M$ and $${\rm D}_vf(x)={\rm D}\tilde f(x)(h-w)\tag5.$$
Turning to the situation of this question, assume $k=d-1$ and let $\Omega$ be an $M$-open neighborhood of $x$ which is $C^1$-diffeomorphic to $\mathbb H^k$. Then, $$w=\langle h,\nu_\Omega(x)\rangle\nu_\Omega(x)\tag6$$ and hence $${\rm D}_vf(x)={\rm D}\tilde f(x)h-\langle h,\nu_\Omega(x)\rangle{\rm D}\tilde f(x)\nu_\Omega(x)\tag7.$$ If $E=\mathbb R$, then we may rewrite this as $$\nabla_vf(x)=\nabla\tilde f(x)-\left\langle\nabla\tilde f(x),\nu_\Omega(x)\right\rangle\nu_\Omega(x)\tag8.$$ This is precisely the definition of the tangential gradient and by the former general reasoning we know that it is independent of the choice of $\tilde f$.