Definition of the Tangent Space I'm watching a series of lectures on differential geometry, and I've run into a bit of a problem with the definition of the tangent space. We first defined a tangent space as $\{(p,v) | v \in \mathbb{R}^n\}$, which makes sense to me: it's the set of all vectors attached at point $p$. We then defined the directional derivative as
$$
(Df)(p,v) = \lim_{t \rightarrow 0} \frac{f(p + tv) - f(p)}{t}
$$
We expanded that to this:
$$
(Df)(p,v) = \left( \sum_{i = 0}^{n}v_i \left.\frac{\partial}{\partial x_i}\right|_{p}
\right) f$$
This makes sense to me; we have defined the directional derivative as an operator that is applied to the function.
Here's the part where I lose the plot. I'm then told that, if I think about it, the portion inside the parentheses is really interchangeable with $(p,v)$. I'm afraid that I've thought about it, and I can't see the equivalence. $\sum_{i = 0}^{n}v_i \left.\frac{\partial}{\partial x_i}\right|_{p}$ is an operator (isn't it?) whereas $(p,v)$ is a ordered pair of elements of $\mathbb{R}^n$. Does that mean that the expression $(p,v)(f)$ makes sense? What would that mean?
I must be thinking about this the wrong way; can someone clarify?
 A: I think the phrase "the portion inside the parentheses is really interchangeable with $(p,v)$" is quite misleading. This is not really true -- as you correctly observed, $(p,v)$ is an ordered pair of elements of $\mathbb R^n$, while the expression in parentheses is an operator on functions.
What is true is that there is a linear map from the set $\{(p,v)|v\in\mathbb R^n\}$ (let's call that the geometric tangent space) into the set of linear differential operators on functions, which takes the pair $(p,v)$ to the "directional derivative operator" that you wrote down. The image of this map is the set of derivations at $\boldsymbol p$, which is the set of all linear maps $X\colon C^\infty(\mathbb R^n)\to \mathbb R$ that satisfy this product rule:
$$
X(fg) = f(p)X(g) + g(p)X(f).
$$
The geometric tangent space is thus canonically isomorphic to the set of derivations at $p$. 
Why does this matter? Because on an abstract manifold, the geometric tangent space doesn't have any coordinate-independent meaning, but the space of derivations at $p$ does. So we take the space of derivations at $p$ as our definition of the tangent space to $M$ at $p$. 
Once you get comfortable with the canonical isomorphism between the geometric tangent space and the space of derivations at $p$, then you might start thinking of them as "interchangeable." But when you're first trying to learn this stuff, it's more productive to think of them as canonically isomorphic.
A: The usual notation for the tangent space at a point $p$ of a differentiable manifold $M$ is $T_pM$. By the definition you can see that this space is  a vector space that has the same dimension $n$ as the manifold $M$. The elements of $T_pM$ are not ''all vectors attached at point $p$'' as you say, but the vectors, attached at $p$, that stay in the tangent plane to the manifold at $p$.
So, in a notation as $ T_pM=\{(p,v):v\in \mathbb{R}^n\}$ $p$ really is simply a parameter, that specifies the point where we take the tangent plane.
The vectors in $T_pM$ can be represented, as you notice, as
$$
\sum_{i = 0}^{n}v_i \left.\frac{\partial}{\partial x_i}\right|_{p}
$$
and we can can think at $T_pM$ as a vector space of linear operators with basis
$$
\left(\frac{\partial}{\partial x_i}|_{p}\right)  \qquad i=1,2\cdots n
 \quad p \in M
$$
We have one such space for every $p$ and we can consider the set of all such tangent spaces:
$$
TM=\bigcup_{p\in M}\{(p,v):v\in T_pM \}
$$ 
In this notation the presence of $p$ has clearly  the sense of parametrize all the tangent spaces, but note that  the element of $T_pM$ is $v$, not the couple $(p,v)$, and $TM$ is the tangent bundle of $M$.
