Definition of ODE and flow on manifold. I have the following confusion in my recent lectures in Riemannian geometry. The idea is to define the notion of Lie derivative using the exponential map. In my lecture notes is the following:
Let $ M $ be a smooth manifold. $ X $ be a vector field on $M$. Let $ F: V \rightarrow U \subset M $ be a local coordinate chart of $ M $ where $ V \subseteq \mathbb{R}^{n} $ is an open set in $ \mathbb{R}^{n} $. If $ u = (u_1,...u_n) \in V $ is the coordinate in $V$. Then we consider the following system of ordinary differential equations:
$$\begin{cases}
du/dt = X(u)\\
\\
u(0)= F^{-1}(p)
\end{cases}$$
for $ p \in M $. The first question I have is what does the above equality means. More precisely, we think of tangent vectors in $ T_{p}M $ as elements that are behaves like directional derivatives. Hence $ X $ should admits the expression $ X = \sum_{j=1}^{n} X_{j} \frac{\partial_{}}{\partial{u_j}} $ which takes values in $ M $ and where $ ( \frac{\partial}{\partial{u_j}} )_{p}$ is defined such that 
$$ (\frac{\partial }{\partial{u_j}} )_p f = \partial_{e_j}( f \circ F \circ F^{-1} \circ p ) $$
where $f$ is a smooth function on $M$ defined locally near $p$. In the lecture, however, it was just said that we should interpret the above equality as
$$ du_j/dt = X_j .$$
My first question is then, is this a consequence of the above discussion? Or simply by definition? After all, $X$ is not meant to take values in $\mathbb{R}^{n}$.
We then defined the solution of the above 
$$e^{tX}(p) = u(t,p)$$
corresponding to the initial position $F^{-1}(p)$. And somehow this becomes an element of $ M $. But following the above discussion $u(t,p)$ is clearly in $ \mathbb{R}^{n} $! The only reasonable explanation I can think of is that $e^{tX}(p) = F( u(t,p) ) $ would be correct notation. This is my second question.
Could somebody explain what the actual definition is? Thanks!
 A: You are strangely mixing objects on the manifold and their coordinate representations. As you stated, $X$ is a vector field in the manifold, it assigns each point $x$ a vector $X(x)$ in the tangent space $T_xM$.
Derivatives of curves in the manifold are geometric objects themselves. Which means that while their arithmetic-analytic definition uses a chart to define the limit of the difference quotient, the resulting vector in the tangent space is independent of the chart.
Thus your differential equation should be $\dot u(t)=X(u(t))$, $u(0)=p$ for a path $u$ in the manifold.
Alternatively, if $u$ is really the parametrization of the path in some chart $F:\Bbb R^n\supset {\cal D}\to M$, then you need to pull back the vector field, $\frac{du}{dt}(t)=F^*X(F(u(t)))$ with initial point $u(0)=F^{-1}(p)$.
What your described problem seems to be is that the letter $X$ is reused for the coordinate vector of the vector field, which of course is different from the vector field on the manifold. It would have been better to use some other letter so that $\xi(p)=F^*X(F(p))$ for the coordinate vector.
There is two insights that should result from your lecture: The local solution theory for ODE on a manifold can be pulled back to a chart where it reduces to the theory of ODE systems in $\Bbb R^n$, so that nothing new or unexpected happens. And second, the local solutions are independent of the chart, so that you get local solutions on the manifold as geometric objects that can be joined to global solutions. In these global solutions you can get, depending on the geometry of the manifold, situations that do not occur in the flat case of  $\Bbb R^n$.
