How Every smooth vector field generates a smooth flow?
Definitions
Smooth Vector Field: A vector field on a smooth manifold $M^n$ is a function $\xi$ on $M^n$, such that $\xi(p)\in T_p(M)\ \forall p\in M^n$ and which is smooth in the following sense: Given local coordinates $x_1,...,x_n$ near $p\in M$, we can write $$\xi(p)=\sum\limits_{i=1}^{n} a_i(p)\frac{\partial}{\partial x_i}$$ and smoothness of $\xi$ means that the $a_i$ are smooth functions.
Smooth Flow: A smooth flow on a smooth manifold $M^n$ is a smooth map $\theta: \mathbb{R} \times M\to M$ such that: (i) $\theta(0,x)=x\ \forall x\in M$ and (ii) $\theta(s+t,x)=\theta(s,\theta(t,x))\ \forall x\in M$ and $s,t\in \mathbb{R}$.
In many places I saw they are constructing a system of ODE's and saying the solution of the system is the required smooth flow. But I didnot understand how it is actually happening?
Fix $p \in M$. Consider the linear first order differential equation : \begin{align} \forall t \in I, ~ \gamma'(t) = X(\gamma(t)),~ \gamma(0) = p \end{align} In charts, this is an ordinary differential equation in $\mathbb{R}^n$ and it has a unique maximal solution, say $\gamma_p : I_p \to M$, with $I_p$ open.
Now, let $U = \bigcup_{p \in M} I_p\times \{p\}$. This is an open subset of $\mathbb{R}\times M$ containing $\{0\}\times M$ (this is the difficult statement of the theorem, I will not provide a proof here). Define on $U$: \begin{align} \varphi^X : U &\to M \\ (t,p) & \mapsto \gamma_p(t) \end{align} One can show that it is a flow and that it generates the vector field $X$.
First, it is a flow because of Cauchy-Lipschitz theorem on uniqueness of solution of differential equation. For $t,s$ such that $\varphi^X(t+s,p)$ and $\varphi^X(t,\varphi^X(s,p))$ is defined, one can show, differentiating with the $t$ variable, that they both are solutions of the same first order ODE with initial data $\varphi^X(s,p)$ at $t=0$, and consequently they are equal.
Moreover, by the very definition of $\varphi^X$: \begin{align} \left.\frac{\mathrm{d}}{\mathrm{d}t}\varphi^X(t,p)\right|_{t=0} = \left.\frac{\mathrm{d}}{\mathrm{d}t} \gamma_p(t)\right|_{t=0}=\gamma_p'(0) = X(\gamma_p(0))=X(p) \end{align} and $\varphi^X$ generates $X$.