Misuse of Tangent Vector I am quite confused with the term tangent used in differential geometry books. It seems to me that people use this word quite loosely. For example, one definition about tangent space in my book is as follows.
The tangent space $T_p(\mathbb R^n)$ at $p\in\mathbb R^n$ is the vector space of all arrows emanating from $p$.
If this is the case, then the tangent space at $p$ can be a vector pointing to any direction. However, intuitively a tangent vector is only meaningful with respect to some kind of geometric object such as a surface or curve where the tangent vector at a point is the vector that only touch the surface or curve at point $p$ if it is extended to a plane or line. My question is how to interpret the word tangent in differential geometry. Is it still consistent with the geometric intuition I have from curve and surfaces, please? Thank you!
 A: In general you define the tangent space to a point $p\in \mathbb R^n$ as the set $$T_p\mathbb R^n=\{(p, v): v\in \mathbb R^n\}=\{p\}\times \mathbb R^n.$$ Notice you may think as the tangent space to $\mathbb R^n$ at $p$ as a copy of $\mathbb R^n$ lying on $p$. Each $T_p\mathbb R^n$ is a vector space with the operations: $$\alpha (p, u)+(p, v)=(p, \alpha u+v).$$ 
With this it is clear $\{(e_1, p), \ldots, (e_n, p)\}$ is a basis for every $T_p\mathbb R^n$ so $T_p\mathbb R^n\simeq \mathbb R^n$ for each $p\in\mathbb R^n$. 
Important: If $p\neq q$ you can't operate $v_p\in T_p\mathbb R^n$ with $v_q\in T_q\mathbb R^n$ for it wouldn't yield a well defined operation. 
Now let us consider the tangent space to a surface embedded in $\mathbb R^n$. Let $M\subset \mathbb R^n$ be a surface. We say $v\in \mathbb R^n$ is a tangent vector to $M$ at $p\in M$ if there exists a differentiable curve $$\gamma:(-\varepsilon, \varepsilon)\longrightarrow M$$ such that $\gamma(0)=p$ and $\gamma^{'}(0)=v$. Again $T_pM$ denotes the set of all tangent vector to $p\in M$. 
This above definition agrees with our intuition.
Obs: Again you can show $T_pM$ is a vector space with the natural operations isomorphic to $\mathbb R^{\textrm{dim}(M)}$.
Exercise: Show that $T_pS^n=\{v\in \mathbb R^n: \langle v, p\rangle=0\}$ (as expected).
