If $P$ is a tangential projection onto the tangent plane $T_\Gamma$ of a surface $\Gamma$ embedded in $\mathbb{R}^d$, defined for $x \in \Gamma$ by \begin{align} P(x) = I - n(x) \otimes n(x) \end{align} where $n$ is the unit normal on $x$. Then it gives the tangential gradient of any vector field $v: \Gamma \to \mathbb{R}^d$ that \begin{align} \nabla_\Gamma v = P \nabla v \end{align}
My question is quite simple: Should we say the tangential projection $P: \Gamma \to T_\Gamma$, or $P: V(\Gamma) \to V(T_\Gamma)$ for a function space $V$?
The definition looks like $P: \Gamma \ni x \mapsto x \in T_\Gamma$, while the tangential gradient acts on a vector field that is a function.
$P$ takes elements of the tangent bundle $T\Bbb R^d$ to elements of the tangent bundle of $\Gamma$. But it only applies to elements of $T_x\Bbb R^d$, where $x \in \Gamma$.
My own choice here would be to name the name the inclusion map $$ i : \Gamma \to \Bbb R^d : x \mapsto x. $$ Then let $$ \zeta = i^{*} (T\Bbb R^d) $$ be the restriction of the tangent bundle of $\Bbb R^d$ to $\Gamma$. Now $P$ is a bundle map from $\zeta$ to $T\Gamma$: $$ P : \zeta \to T\Gamma. $$
To answer what you asked: $P$ is definitely not a map from $\Gamma$ to $T\Gamma$, for the argument to $P$ is a vector tangent to $\Gamma$, not a point of euclidean space.