You can prove it a number of ways so here's a simple-ish one :
Take Minkowski space $M$, with the manifold atlas $(\mathbb{R}^n, \operatorname{Id})$. I think it's no big secret that, for cartesian coordinates, the geodesic equation is
$$\ddot{x}^\mu(\tau) = 0$$
with solution
$$x^\mu(\tau) = x_0^\mu + v_0^\mu \tau$$
If we consider the tangent space at a point $p$ with coordinates $x_0$, any geodesic starting at $p$ and with tangent vector $v^\mu_0$ has this form. What we want is a map from every point of $T_pM$ to $M$ from there. If we take any point $q \in M$ with coordinates $y^\mu$, we have that
$$x^\mu(1) = y^\mu$$
for some $v_0 \in T_pM$. It can be shown fairly easily by considering, component-wise
$$v_0^\mu = y^\mu - x_0^\mu$$
There is a map from some subset of $T_p M$ to $M$. It's not too hard to show the reverse and that overall, this maps $T_p M$ to $M$, giving you some map
$$\phi : T_pM \to M$$
If you also add the Minkowski metric on $M$ as a metric space, we may want to check that $g_p(v,v) = \langle \phi(v), \phi(v) \rangle$, which is just
$$\eta_{\mu\nu} v^\mu v^\nu = (v^0)^2 - \vec{v} \cdot \vec{v}$$
and
$$\langle \phi(v), \phi(v) \rangle = (x_0^0 - y^0)^2 - (\vec{x}_0 - \vec{y}) \cdot (\vec{x}_0 - \vec{y})$$
Which are the same value indeed.
