Geometric formulation of the equivalence principle Let $(M,g)$ be a $4$-dimensional Lorentzian manifold. It is well know that given $(U,\psi=(x^1,\ldots,x^4))$  local chart around some $p\in M$, it is posible to find a change of coordinates given by $(U,\varphi=(\bar x^1,\ldots,\bar x^4))$ such that the components $\bar g_{ij}$ of $g$ respect $\{d\bar x^i\otimes d\bar x^j\}_{i,j}$ are equal to the components of $\eta$ (minkowskian metric on $\mathbb{R}^4$) at $p\in U$, and also the partial derivatives $\partial_{x^k}g_{ij}(p)$ vanishes. [See Differential geometry and relativity theory, by Richard L. Faber, pag 178]
In this sense, can we stablish the equivalence principle as follow?
The space-time is a $4$-dimesional Lorentzian-manifold.
 A: Actually the result is even stronger: 
Given a timelike geodesic $\gamma$ and a point $p \in \gamma$, there is a neighborhood $U \ni p$ equipped with coordinates, $x^0,x^1,x^2,x^3$ such that in the portion of $\gamma$ included in $U$, exactly along $\gamma$, the derivatives of the metric vanish in the said coordinates.
Equivalently the Christoffel symbols $\Gamma^a_{bc}$ in the said coordinates vanish along $\gamma$ in $U$.
The coordinate  $x^0$  coincides with the proper time measured along $\gamma$ and the remaining three coordinates $x^k$ can be chosen spacelike and orthogonal to $\gamma$. 
The mentioned coordinates are called Fermi coordinates adapted to $\gamma$
This result (but also the weaker one you mention does since in the proof below we use the fact that the Christoffel symbols vanish exactly as the origin of the coordinates) implies a geometric version of the equivalence principle. More precisely it entails the statement saying that, 
in the reference frame centered on a free falling body, the motion of another free falling body is approximated by a constant velocity motion and this approximation is valid for short times and in a little spatial region around the center of the free falling reference frame.
Let us illustrate how it happens. Consider the said coordinate system $x^0,x^1,x^2,x^3$ supposing (by redefining the origin of coordinates if necessary) that the portion of $\gamma$ in $U$ is described by $x^0 \in (-a,a)$ and $x^k=0$ for $k=1,2,3$.
A second timelike geodesic $\gamma'$ crossing $\gamma$ at the origin has equation
$$\frac{\mathrm d^2x^a}{\mathrm dt^2} = -\Gamma^{a}_{bc}\frac{\mathrm dx^b}{\mathrm dt}\frac{\mathrm dx^c}{\mathrm dt}\:.$$
However, exactly at the origin of the coordinates where the stories of the two free falling bodies meet and  supposing to take the proper time of $\gamma'$ to be  $t=0$ there,
$$\frac{\mathrm d^2x^a}{\mathrm dt^2}\bigg|_{t=0} = -
\Gamma^{a}_{bc}\bigg|_{(0,0,0,0)}
\frac{\mathrm dx^b}{\mathrm dt}\bigg|_0
\frac{\mathrm dx^c}{\mathrm dt}\bigg|_0 = 0 
~~\frac{\mathrm dx^b}{\mathrm dt}\bigg|_0\frac{\mathrm dx^c}{\mathrm dt}\bigg|_0 =0\:.$$
Expanding the expression of $\gamma'$ in coordinates around $t=0$,
$$x^a(t) = x^a(0) + \frac{\mathrm dx^a}{\mathrm dt}\bigg|_0 t + \frac{1}{2}\frac{\mathrm d^2x^a}{\mathrm dt^2}\bigg|_0 t^2 + O^a(t^3)= 0 + V^a t + 0 + O^a(t^3)$$ that is
$$ x^a(t)  = V^at + O(t^3)\:.$$
This is in fact  a motion with constant velocity. Notice that $V^0 \neq 0$ because $\gamma'$ is timelike, so that we can re-parametrize the geodesic using the time of the coordinates $x^0$ instead of the proper time $t$ along $\gamma'$. Defining $v^k := V^k/V^0$ for $k=1,2,3$ we easily have
$$x^k(x^0) = v^k x^0 + O^k((x^0)^3)\:.$$ 
To appreciate some acceleration we have to deal with infinitesimal of the third order  $O((x^0)^3)$ instead of the second order. This approximation is as better as $x^0$ is smaller, i.e., the body with story given by $\gamma'$ is close to $\gamma$. 
