Differentiation on Manifolds Basics I'm having some real trouble comprehending integral curves and Lie derivatives on a Manifold. I will write out my understanding and ask the questions below. 
For a vector field $X$ on smooth manifold $M$ there is a unique curve $\gamma (t):I\rightarrow M$ through each point $p\in M $ such that $\gamma (0)=p$ and such that the tangent vector to the curve $\gamma$ at $\gamma (t)$ is $X_{\gamma (t)}$. This is the integral curve of $X$ through $p$. (apparently) we view $I$ as a manifold (why, what is $I$ precisely?) with coordinates $t$ then $d/dt$ is a tangent vector field on $I$ which acts on functions $f:I\rightarrow \mathbb R$. By definition the tangent vector to the curve is given by $\gamma '(t):= \gamma _* (d/dt)_{\gamma (t)}=X_{\gamma(t)}$. 
For local coordinates about $p\in M$
\begin{equation}
X_{\gamma(t)}(x^i)=\frac{d}{dt}\gamma ^*(x^i)=\frac{d}{dt}(x^i\circ \gamma)=\frac{d}{dt}\gamma^i(t)=\frac{d}{dt}x^i(t)
\end{equation}using $X=X^j\partial /\partial x^j$ then,
\begin{equation}
X_{\gamma(t)}(x^i)=X^j\frac{\partial x^i}{\partial x^j}\bigg|_{\gamma(t)}=X^i(\gamma (t))=X^i\{x^i(t)\}
\end{equation}
We can equate the final results of each line to get a set of differential equations.
\begin{equation}
\frac{d}{dt}x^i(t)=X^i\{x^i(t)\}
\end{equation}
We integrate these to get the curve $\gamma$. At each point $p\in M$ we move a parameter a distance $t$ along the integral curve $\gamma$ of $X$ through $p$. Doing this for all points of $M$ we get the flow along $X$. The vector field generates the flow. 
My question is can anyone explain why this is different to a set of tangent vectors in the tangent space to $M$? Is the integral curve a function on $M$? Also It looks as if we are using the pullback of the coordinate system? Thanks for your help !! 
I am hoping that if I understand this I will not need to ask my next question regarding the Lie derivative of a tensor field on $M$. 
 A: So I can try to give you some of my intuition about integral curves. 
Let $M$ be our manifold.
$T_pM$ is the space of all vectors tangent to a point $p \in M$. This is called the tangent space at $p$.
$TM$ is the disjoint union of $T_pM$ over all $p \in M$. This is called the tangent bundle of $M$.
$X(p)$ is the vector field on $M$. This gives us a tangent vector in $T_pM$ for every point $p \in M$. It is instructive to write $X(p) \in T_pM$.
$\gamma(t)$ is our integral curve. This uses some real number $t$ as a way to keep track of where we are when we look at the curve on the manifold $M$. It effectively takes in a real number and pumps out a point $p \in M$.
The important feature about integral curves is that they always follow the "flow" of your vector field. So we can think of this as $\gamma(t)=p \in M$. Now we look at the tangent space at $p$, which is given by $T_pM$, and then pick out the tangent vector that tells $\gamma(t)$ how to "flow". This is given by $X(p)=X(\gamma(t)) \in T_{\gamma(t)}M = T_pM$.
We then see that $X(p)$ is just the tangent vector at each point along the integral curve. This can be written as $X(\gamma(t))=\gamma'(t)$. Now we have an explicit relationship between the vector field and how it determines the path that the integral curve takes across the manifold.
So to answer your questions:
Q1) Why do we use $I$?
A1) The interval $I$ is just a simple way to parameterize a curve when we have start and stop points. We could use any parameterization really, as it is mostly a matter of convenience. If you don't have start or stop points, then you could just use an open interval of reals from $a$ to $b$. Most of the time I think you would use an open interval though if you wanted to parameterize an infinitely long integral curve (you just make sure $t=0$ gives you some point you want the curve to pass through on the manifold).
Q2) Why is this* different than a set of tangent vectors in the tangent space to $M$?
A2) $X(\gamma(t))=\gamma'(t)$ indicates that the derivative of the integral function is a subset of tangent vectors in the tangent space $TM$. So it is pretty much a parameterized vector. 
Q3) Is the integral curve a function on $M$?
A3) Technically it is a function from your parameter $t$ (typically the reals) to some $p \in M$. Intuitively, it is a curve living on the manifold.
Hope that helps.
