Express the flow of $X + Y$ in terms of flow of $X$ and flow of $Y$ Let $X, Y$ be two smooth vector fields on some smooth manifold $M$, and let $F(t,p), G(t,p)$ be the flow for $X, Y$ respectively. Then is it possible to express $X+Y$, which is another smooth vector field on $M$, in terms of $F$ and $G$?
For simplicity assume we have a global flow. There are two conditions for a map $F:\mathbb{R}\times M \rightarrow M$ to be the flow of X, one is that $\forall p \in M, F(0,p) = p$; the other is that $\forall p \in M, \frac{\partial}{\partial t}|_{(0,p)}F=X|_p$.
I was considering $F+G$ as the flow, since $\frac{\partial}{\partial t}|_{(0,p)}(F+G)=X|_p+Y|_p$. But then $(F+G)(0,p) = p+p$ which is not necessarily defined in an arbitrary manifold. I also considered composing $F$ and $G$, but $F \circ G \neq G \circ F$ in general, so it doesn't seem correct.
Or to ask it in a more general way, suppose we have a coordinate chart $(U, \phi=(x^1, ..., x^n))$ and a local flow $F:(-\varepsilon, \varepsilon) \times W \rightarrow U$ of some vector field $X = \sum_{i=1}^{n}a^i \frac{d}{dx^i}$ on $U$, then how do we express $F$ in terms of the flow for each individual $\frac{d}{dx^i}$?
 A: Thanks to the help of one of the TA in my differential geometry course, I can answer my question now!
Claim: when $X, Y$ Lie commute, the flow of $X+Y$ is just $F_t \circ G_t (p)$ (or $G_t \circ F_t(p)$ since the two commutes).
Proof:
Clearly $F_0 \circ G_0 (p) = p$. Now suppose the vector fields Lie commute, that is $[X, Y] = 0$. This is equivalent to $Y$ invariant under $X$, that is $F_{t*,p}(Y|_p) = Y|_{F_t(p)}$. Then
$$\frac{d}{dt}|_0 (F_t \circ G_t) (p) = \frac{d}{dt}|_0 F(t, G_t(p)) = F_{0*,G_0(p)}(\frac{d}{dt}|_0) + F_{0*,G_0(p)}(G_{0*,p}(\frac{d}{dt}|_0)) = X|_p+F_{0*,p}(Y|_p)=X|_p+Y|_p$$
where $F_{*,p}$ stands for the differential of $F$ at $p$.
Therefore the flow is as desired. $\blacksquare$
If, on the other hand, $X$ and $Y$ do not Lie commute, we see that $Y$ is no longer invariant under $X$, then it becomes much more complicated and the solution could be found here https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula.
If there is anywhere wrong or unclear about my proof, please let me know!
