One-to-one correspondence between the flow of an autonomous ODE an its solutions I want to show the following that there is a bijective correspondence between a certain group action $g$ on $\mathbb{R}$ (I believe that that's called a flow - I don't know much about ODE's, so please keep the explanations at an accessible level) and global solutions of a certain type of differential equations (so talking about $g$ is the same as talking about solving certain ODE's), i.e. I want to show the following two theorems:
"1) Let $g:\mathbb{R}^{n}\times\mathbb{R}\rightarrow\mathbb{R}^{n}$
be a group action of the additive group $\mathbb{R}$ on $\mathbb{R}^{n}$
such that $\partial_{2}g$ exists on the whole domain. Then there exists the ODE 
$$
x'=\partial_{2}g\,\left(x,0\right)\left(x\right),
$$
such that for a solution $l:\mathbb{R}\rightarrow\mathbb{R}^{n}$ with initial value $x_{0}\in\mathbb{R}^{n}$of the above, we have: $l$ is unique and has the form $l\left(t\right):=g\left(x_{0},t\right)$.
(Note, that $\partial_{2}g$  denotes the vector of the partial derivates
with respect to the time/the $n+1$'th argument of $\phi$).
2) Conversely if an ODE $$
x'=f\left(x\right)
$$ has unique global solutions, then there exists a mapping $g:\mathbb{R}^{n}\times\mathbb{R}\rightarrow\mathbb{R}^{n}$, such that $g$ is a group action ( like described above) and $\partial_{2}g\,\left(x,0\right)=f\left(x\right)$ holds."
My questions are: 1) Have I even stated this proposition correctly ? Is it true what I'm asking ?
2) How can I prove that ? For 1) for example it is easy to show (thanks to the properties
of the group action) that if $l$ has the above form,
that it is a solution. But taking an arbitrary solution and proving
that $l$ has to be of the above form seems to be very difficult.
 A: I'm not sure I fully understand your question.  Let me rephrase it differently and you can tell me if my rephrasing is actually what you want.  To avoid having to think about things that are too pathological, I will assume your $\mathbb R$ action has a continuous time derivative. 
Before rephrasing anything, we notice that your condition on the vector fields $f(x)$ in the ODE $\dot x = f(x)$ means that their flows generate a smooth, additive $\mathbb R$ action.  Do you want a proof of this claim?  It follows from the local existence theorem, coupled with your hypotheses that you have global existence and uniqueness of solutions to the initial value problem and that $f(x)$ does not depend on $t$.  (The jargon: $t$ independent vector fields give rise to autonomous differential equations.)
Now I can reformulate what I understand your question to be.  Let $\mathcal F$ be the set of differentiable additive $\mathbb R$ actions on $\mathbb R^n$.  Let $\mathcal X$ be the set of vector fields on $\mathbb R^n$ that generate flows with the properties you list (we can obtain ``most'' of these by requiring that the vector field be locally Lipschitz and satisfy some growth condition, but this isn't directly relevant).
You have an obvious map $\mathcal X \to \mathcal F$.  This map takes a vector field and spits out the flow that comes from it.  Your questions are (1) is this map surjective and (2) is this map injective?  
Let's start with (2): the map is injective.  This follows from the fact that we can recover the vector field by observing that if $g(x,t)$ is your flow (i.e. $\frac{d}{dt} g(x, t) = f( g(x,t) )$, $g(x, 0) = x$), then $\frac{d}{dt}|_{t=0} g(x,t) = f(x)$, recovers the vector field.  
The map fails to be surjective, however, because there are ODEs that almost generate flows while not satisfying the global uniqueness property.  We will notice first that any smooth, additive $\mathbb R$ action comes from an ODE in a sense.  Define $F(x) = \partial_2 g(x, 0)$.  This is the candidate for being our vector field.  The additive property gives 
\[
g(x, t+s) = g( g(x, s), t)
\]
Differentiate both sides with respect to $t$ and evaluate at $t=0$.  This gives
\[
\partial_2 g(x,s) = (\partial_2 g)(g(x,s), 0) = F(g(x,s)).
\]
We already have that $g(x,0) = x$, so this indeed looks like a flow generated by the ODE $\dot x = F( x )$.  
This doesn't actually show the map is surjective, however... remember that you wanted your vector fields to have global uniqueness.  Consider this example on $\mathbb R$:
[
\dot x = 3 x^{2/3}.
]
The ``flow'' it wants to generate is the $\mathbb R$ action 
\[
g(x, t) = (t + x^{1/3} )^3.
\]
Notice that this $g(x,t)$ defines a flow in the sense you wanted.  However, the ODE I wrote down above does not satisfy uniqueness.  (There are infinitely many solutions satisfying the IC $x(0) = 0$... notably $x(t) = t^3$ and $x(t) = 0$.)  Thus, this isn't in the image of the map $\mathcal X \to \mathcal F$.
