When is a flow geodesic and how to construct the connection from it Let $(M,\Gamma)$ be a $C^\infty$ $n$ dimensional real manifold with a linear connection $\Gamma$ on it. I know the following:
If $\gamma:[t_0,t_1]\rightarrow M$ is a smooth curve and is a geodesic, then in local coordinates we have $$ \ddot{\gamma}^\mu(t)+\Gamma^\mu_{\alpha\beta}(\gamma(t))\dot{\gamma}^\alpha(t)\dot{\gamma}^\beta(t)=0. $$ By defining $\dot{\gamma}(t)=v(t)$, this is a first-order system of ODEs: $$ \frac{dv^\mu}{dt}=-\Gamma^\mu_{\alpha\beta}v^\alpha v^\beta \\ \frac{d\gamma^\mu}{dt}=v^\mu. $$ Since this is an ODE for the variables $(x^1,...,x^n,v^1,...,v^n)$, this is a differential equation on the tangent bundle of $M$, $TM$, moreover, this is a first-order, homogenous differential equation on $TM$, so it is represented by a vector field, $G\in\Gamma(TTM)$, which I'll call the geodesic flow.
I also know that this vector field can be defined invariantly if we take $\gamma_{(p,v)}(t)=\exp_p(tv)$, eg. the maximal geodesic with initial point $\gamma_{(p,v)}(0)=p$ and $\dot{\gamma}_{(p,v)}(0)=v$ and we take $\bar{\gamma}_{p,v}(t)=(\gamma_{(p,v)}(t),\dot{\gamma}_{(p,v)}(t))$ the natural lift of the curve to $TM$, then $$G_{(p,v)}=\left.\frac{d}{dt}\bar{\gamma}_{(p,v)}\right|_{t=0}.$$ In local coordinates $(x,v)$ this obviously has the form $$ G_{(p,v)}=v^\mu\left.\frac{\partial}{\partial x^\mu}\right|_{p}-\Gamma^\mu_{\alpha\beta}(p)v^\alpha v^\beta\left.\frac{\partial}{\partial v^\mu}\right|_{v}. $$
I don't have any textbooks that deal with this any further however, so I have two questions.
Question 1: Given a smooth vector field $G\in\Gamma(TTM)$, what is the criteria for $G$ to be the geodesic flow of a linear connection? I can see from the local coordinate formula that the $x^\mu$ components contain only $v^\mu$ and the $v^\mu$ components are a quadratic form of the $v$ coordinates, but I am looking for invariant characterization. It is also clear to me that $G$ must not be vertical anywhere, but I don't think this is enough.
Question 2: If we are given $G$, how can I recover the torsionless connection? I mean, if I write up the local coordinate form, I can read off the connection coefficients and define the covariant derivative as $\partial_\mu V^\nu+\Gamma^\nu_{\mu\sigma}V^\sigma$ but I am once again looking for invariant characterization. A limit or $t$-derivative or something like that which reproduces the covariant derivative on the base if I know $G$.
 A: Here is a geometric way that turns out to be equivalent to Robert's answer (i.e., to the Klein-Grifone-Foulon approach to connections associated to a second order ordinary differential equation of a manifold). 
Let $\phi_t : TM\setminus 0 \rightarrow TM\setminus 0$ be the (local) flow of the second order equation, and let $D\phi_t : T(TM\setminus 0) \rightarrow T(TM\setminus 0)$ be its differential, which is also a flow. 
In order to define a complement to the vertical space $V_{v_x}$ at the point $v_x \in  TM\setminus 0$, consider the three subspaces 
$L(0) = V_{v_x}$, $L(t) := D\phi_{-t}V_{\phi_t(v_x)}$, and $L(2t) := D\phi_{-2t}V_{\phi_{2t}(v_x)}.$
These are three $n = \dim(M)$ dimensional subspaces in the $2n$-dimensional vector space $T_{v_x}(TM \setminus 0)$ and they are pairwise transversal if $t$ is close to zero.  Now consider the harmonic conjugate (i.e., as in standard projective geometry) of this triplet: the image of $L(t)$ under the reflection that is minus the indentity on $L(0)$ and the identity on $L(2t)$. As $t$ tends to zero this harmonic conjugate gives you the complement to the vertical space (i.e., the horizonal subspace of the connection). 
A: Note:  I've decided that this answer should be rearranged a bit
so that it clearly separates the discussion of the basic properties of 
the tangent bundle from the discussion of the formulae associated to a connection.
The content is the same, but I hope it's clearer.
The standard way to discuss the geometry of connections and geodesic flow 'invariantly' (by which, I assume you mean 'without reference to coordinates') is to exploit the underlying geometric features of the tangent bundle $\pi:TM\to M$.  There are three that are important here:
First, because $TM$ is a vector bundle over $M$, there is a flow associated with
scalar multiplication, $S_t(w) = e^tw$ for $w\in TM$, and this is the flow of 
a unique vector field $R$ on $TM$.
Second, since each diffeomorphism $\phi:M\to M$ canonically induces a diffeomorphism $\phi':TM\to TM$, each differentiable vector field $X$
on $M$ induces a vector field $X'$ on $TM$, known as the tangential prolongation of $X$. The assignment $X\mapsto X'$ is a Lie algebra homomorphism 
and $X$ is $\pi$-related to $X'$.
Third, because $T_{\pi(w)}M\subset TM$ is a vector space, there is a canonical
isomorphism $\iota_w:T_{\pi(w)}M\to T_w(T_{\pi(w)}M)$ for each $w\in TM$.  Set
$$
\nu_w = \iota_w\circ\pi'(w): T_w(TM)\to T_w(T_{\pi(w)}M)\subset T_w(TM). 
$$ 
Then $\nu:T(TM)\to T(TM)$ is a canonically defined nilpotent linear endomorphism 
of the vector bundle $T(TM)$ whose kernel and image are $\ker(\pi')=V\subset T(TM)$,
i.e., the so-called vertical bundle.
Now consider a torsion-free linear connection $\Gamma$ on $TM$ and its
associated geodesic flow vector field $G$.  
First of all, one can verify (using the formulae given in the question and the
above definitions) that $G$ satisfies the two conditions
$$
\nu(G) = R\qquad\text{and}\qquad [R,G] = G.
$$ 
(As usual, $[,]$ denotes the Lie bracket of vector fields on $TM$.)
Conversely, by Euler's Theorem, if a smooth vector field $G$ on $TM$ 
satisfies these two conditions, then it is the geodesic flow vector field 
of a torsion-free affine connection $\Gamma$ on $TM$, as the coordinate
formulae show.  
One can go further and see how $G$ and the connection define each other
without using coordinates:
Recall that a connection $\Gamma$ on $TM$ can be thought of as a choice 
of a splitting $T(TM) = V \oplus H$, where $V=\ker (\pi')\subset TM$ 
and $H$ is the so-called 'horizontal space' of the connection, 
so that $\pi'(w):H_w\to T_{\pi(w)}M$ is an isomorphism for each $w\in TM$.
In particular, every vector field $X$ on $M$ has a unique horizontal lift $X^H$,
which is the unique section of $H$ on $TM$ that is $\pi$-related to $X$.
The canonical map $\nu:T(TM)\to V$ 
restricts to define an isomorphism of bundles $\nu:H\to V$, 
and $G$ is the unique section of $H$ that satisfies $\nu(G) = R$.
This is how $H$ determines $G$.
To recover $H$ from $G$, one has the identity
$$
X' - \bigl[\,G,\,\nu(X')\,\bigr] = 2\,X^H.
$$
Thus, the knowledge of $G$ suffices to determine $X^H$ 
for all vector fields $X$ on $M$, which, of course, determines $H$.
(Note that, because $\nu(X')$ does not depend on derivatives of $X$,
the left hand side of the above equation would be at most first order in $X$,
but, as the above formula shows, it is actually of order $0$ in $X$.)
Even more, an explicit formula for the covariant derivative 
associated to the connection can be written as follows:
Recall that, if $X$ is a differentiable vector field on $M$ and $w\in TM$ is given,
then the connection $\Gamma$ defines a unique element $\nabla_wX\in T_{\pi(w)}M$.
Using the canonical isomorphism $\iota_w:T_{\pi(w)}M\to T_w(T_{\pi(w)}M)$, 
one can thus regard $\nabla_wX$ canonically as an element 
of $T_w(T_{\pi(w)}M) = V_w$ for each $w\in TM$, thereby defining a section
$\nabla X$ of the vertical bundle $V\subset T(TM)$ over $TM$.
One then has the identity 
$$
\nabla X = X' - X^H = \tfrac12\bigl(\,X' + [G,\nu(X')]\,\bigr),
$$
valid for all differentiable vector fields $X$ on $M$.
