Given a vector field $X$ on the manifold $M$ its flow is, loosely speaking, a map $F= F(t,x)= F^t(x)$, in the variables $(t,x)\in I\times M$, such that the curve $t\mapsto F^t(x_0)$ is the unique solution of: $$\frac{d}{dt} \Big|_{t = t_1}F^t(x_0) = X(F^{t_1}(x_0))\qquad \forall x_0,t_1\qquad (1)$$ with initial condition $F^0(x_0)= x_0.$ If this holds, then $F$ can be shown to satisfy the group law $F^{t+s} = F^t\circ F^s$ (it follows from the uniqueness theorem for ODEs).
Question: assume only that $F = F^t(x)$ is some map which satisfies $F^0(x) = x$. Is the following condition $(2)$ equivalent to condition $(1)$? $$\frac{d}{d \epsilon} \Big|_{\epsilon = 0}F^\epsilon(x) = X(x)\qquad \forall x \qquad(2)$$ In other words, is $(2)$ sufficient to show that $F$ is exactly the flow of $X$?
I want to show that $(1)$ holds, so I start writing $$\frac{d}{dt}\Big|_{t=t_1}F^t(x_0) = \frac{d}{d\epsilon}\Big|_{\epsilon=0}F^{t_1+\epsilon}(x_0).$$ Now if the group law was true (which I don't know), then $F^{t_1+\epsilon}(x_0) = F^\epsilon(F^{t_1}(x_0))$ and the conclusion follows immediately from $(2)$. But I don't know if the group law follows from $(2)$, basically because I can't apply directly the uniqueness theorem for ODEs. What am I missing? Are there additional necessary conditions?
Allow me to do a little yoga with Lie groups. You should take this as philosophy, though it can be made into rigorous mathematics.
In a Lie group $G$, a 1-parameter subgroup is a Lie group homomorphism $\gamma: \Bbb R \to G$. The following are equivalent.
(1) If you have a smooth homomorphism $\gamma: \Bbb R \to G$, it is determined by $\gamma'(0)$, by the existence and uniqueness theorem for ODEs. (Precisely, it's the unique curve with $\gamma'(t) = dL_{\gamma(t) }(\gamma'(0))$; hwere $L_g$ is left-multiplication by $g \in G$.) You could call this the flow of the element $\gamma'(0) \in \mathfrak g$.
(2) If you have a smooth curve satisfying $\gamma'(t) = dL_{\gamma(t) }(\gamma'(0))$, it is the flow of $\gamma'(0)$ and in particular is a 1-parameter subgroup.
But there are a great many curves $\gamma: \Bbb R \to G$ with $\gamma(0) = 1_G$ and $\gamma'(0)$ fixed that are not homomorphisms! As the sillest example, take $G = \Bbb R$; there are obviously a lot of functions on $\Bbb R$ with $f(0) = f'(0) = 0$ that are not identically zero. This condition gives us so little information on its own that we could not possibly say much about $f$ - it doesn't just give us information only local to $0$, it only tells us about what the derivative of $f$ looks like there, not any of the higher derivatives.
Now for the part that is harder to make rigorous: consider the Lie group $G = \text{Diff}_c(M)$ of compactly supported diffeomorphisms on a smooth manifold $M$. (A compactly supported diffeomorphism is one that is the identity outside a compact subset of $M$.) This is a kind of infinite-dimensional Lie group; to describe its smooth structure you need to come up with a notion of something called a "Frechet manifold". Setting up this theory is quite challenging. The Lie algebra of $G$ is the space $\mathfrak X_c(M)$ of compactly supported vector fields on $M$ (those that are zero outside a compact set). The above, applied to $G$, becomes
(1) If $F^t$ satisfies the group law and has $\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0} F^\varepsilon = X$, then $F^t$ is the flow of the compactly supported vector field $X$.
(2) If $F^t$ is the flow of $X$, then it satisfies the group law.
But the observation about smooth curves to $G$ still holds: there are a great many smooth curves $\gamma: \Bbb R \to \text{Diff}_c(M)$; these are smooth maps $F^t: \Bbb R \times M \to M$ such that, for fixed $t$, $F^t$ is a compactly supported diffeomorphism of $M$. They do not have to be group homomorphisms, and hence do not have to be the flow of a vector field. Thus your (1) and (2) are quite different; as above, we just do not have information about $F^t$ away from $t=0$, and there, we only know about its derivative, not the higher derivatives in $t$.