What exactly is a tangent vector? I'm a physics student with a very loose understanding of the mathematics I use. I'm trying to learn a little more about very basic topology, manifolds, and Riemannian geometry. I'm using Nakahara's Geometry, Topology, and Physics for self-study. I'm trying to read up about vectors on manifolds and the concept of a tangent vector has me thoroughly confused. Up until now I had always pictured the tangent space something like a plane tangent to a point on the surface of a manifold, however if I'm understanding my book correctly the elements of the tangent space seem to be something more like differential operators. I will lay out the definition from my book and then see if perhaps you guys might be able to help me understand exactly what a tangent vector, and what the tangent space is.
My book defines a tangent vector at $c(0)$ as the direction derivative of a function, $f:M \rightarrow \mathbb{R},$ along a curve $c:(a,b) \rightarrow M,$ or $c(t)$ where $a<0$, $b>0$ at $t=0$. The rate of change is
$$\frac{df(c(t))}{dt} \Bigr|_{t=0}.$$
Assuming we've defined all coordinates correctly and stuff this becomes
$$\frac{\partial f}{\partial x^\mu} \frac{d x^\mu(c(t))}{dt} \Bigr|_{t=0}.$$
Defining the differential operator $X$ as
$$X=X^\mu \left( \frac{\partial}{\partial x^\mu} \right),$$
$$X^\mu=\frac{dx^\mu(c(t))}{dt} \Bigr|_{t=0},$$
the directional derivative can be found by applying $X$ to $f$. Putting it all together, this means
$$\frac{df(c(t))}{dt} \Bigr|_{t=0} = X^\mu \left( \frac{\partial f}{\partial x^\mu} \right) = X[f].$$
The book then says, "the last equality defines $X[f]$. It is $X=X^\mu \partial / \partial x^\mu$ which we now define as the tangent vector to $M$ at $p=c(0)$ along the direction given by the curve. This makes the tangent vector a differential operator, it seems like. Intuitively I had thought that the tangent vector would be $X^\mu \partial f / \partial x^\mu$ along the direction of the curve.
The book goes on to say that instead of identifying a tangent vector with a curve you instead identify a tangent vector with the following equivalence class of curves
$$[c(t)] = \left\{ \tilde{c}(t) \Bigr| \tilde{c}(0)=c(0) \text{ and } \frac{dx^\mu (\tilde{c}(t))}{dt} \Bigr|_{t=0} = \frac{dx^\mu (c(t))}{dt} \Bigr|_{t=0} \right\},$$
and that the tangent space is the collection of all such equivalence classes at $c(0)=p.$ So I'm a little confused. The equation above seems to define a tangent vector as an equivalence class of curves. The previous definition seems to define a tangent vector as a differential operator. My intuition tells me something different altogether. Can someone help me sort through all this? Thanks.
 A: The equivalence relation on the (parametrised) curves is exactly that they give the same differential operator at the point $p$; this reconciles those two definitions.  See below for some more details.
As for intuition, that can be a little harder.  The thing to recognise is that we need a definition of tangent vectors without reference to an ambient space.  In other words, we may have a manifold defined abstractly, rather than embedded in some Euclidean space (in general relativity, this is often the case).
Let $\{y^i\}$ be coordinates on $n$-dimensional Euclidean space.  Notice that if we take a 'vector' $\vec{v} = (v^1, \ldots, v^n)$, we can think of this as giving the derivatives of the coordinate functions along the straight line parametrised by $y^i(t) = y^i_0 + v^i t$.  Indeed, the directional derivative of any function $f$ along this line is given by
$$
\frac{df}{dt} = \sum_i v^i \frac{\partial f}{\partial y^i}
$$
So we can associate $v$ with the differential operator $\sum_i v^i\partial/\partial y^i$.  You can easily check that this gives a linear isomorphism between the vector space $\mathbb{R}^n$ and the space of operators spanned by $\{\partial/\partial y^i\}$.
Let me return briefly to the 'equivalence classes of curves' bit.  Choose some other parametrised curve $\gamma : [0,1] \to \mathbb{R}^n$ passing through the same point, i.e. $y^i(\gamma(0)) = y^i_0$.  Then the directional derivative of $f$ along the curve at the point $(y^1_0,\ldots, y^n_0)$ is
$$
\left.\frac{df(\gamma(t))}{dt}\right\vert_{t=0} = \sum_i \left.\frac{\partial y^i(\gamma(t))}{\partial t}\frac{\partial f}{\partial y^i}\right\vert_{t=0} 
$$
We define two curves $\gamma, \gamma'$ to be equivalent iff they give the same directional derivative for every function $f$, i.e. the same differential operator.  This gives our isomorphism between equivalence classes of parametrised curves passing through some point and first-order differential operators at the point.  Note that rescaling the parameter $t$ will rescale the differential operator, which is why we talk about parametrised curves.
Now suppose we have some embedded manifold $X$, with local coordinates $x^\mu$, and embedding functions $y^i(x^\mu)$.  A tangent vector (in the familiar sense) to $X$ just gives the infinitesimal change in the coordinates $y^i$ when we change the coordinates $x^\mu$ by an arbitrary infinitesimal amount $\delta x^\mu$.  We have
$$
\delta y^i = \sum_\mu \delta x^\mu \frac{\partial y^i}{\partial x^\mu} 
$$
So the corresponding tangent vector, using the notation we introduced before, is
$$
\sum_{i, \mu} \delta x^\mu \frac{\partial y^i}{\partial x^\mu} \frac{\partial}{\partial y^i}
$$
The $\delta x^\mu$ are arbitrary, and as we vary them, we map out a linear subspace spanned by the 'vectors'
$$
\sum_i \frac{\partial y^i}{\partial x^\mu} \frac{\partial}{\partial y^i} = \frac{\partial}{\partial x^\mu}
$$
Now we realise that the operators $\partial/\partial x^\mu$ don't actually depend on the embedding at all, and we have successfully defined tangent vectors in an intrinsic way on any differentiable manifold!
This has become a very long answer, but I hope it's helpful.
