Very briefly. The line of reasoning is the following: the acceleration $A^{\mu}$ is GR is
rather formal construction, it is the covariant derivative of the speed
$U^{\mu}$ with respect to some natural parameter $\lambda$ which parameterize
a trajectory. For massive particles you can choose this parameter to be proper
time $d\tau$ thus $A^{\mu}=DU^{\mu}/d\tau$, although it is not possible for
massless particles, for which $d\tau=0$.
Therefore your question is related to the following one: what is $DU^{\mu}$? It
turns out that the simplest (and only) way to construct the covariant
differential of a vector field is to compare two infinitesimally separated
vectors in the same point (it is essential), e.g., the vector $U^{\mu}\left(
x^{\alpha}+dx^{\alpha}\right) $ and the vector $U^{\mu}\left( x^{\alpha
}\right) $ which should be subject to a parallel translation to the point
$x^{\alpha}+dx^{\alpha}$. After the parallel translation we obtain a new
infinitisimally close vector $U^{\mu\prime}=U^{\mu}\left( x^{\alpha}\right)
+\delta U^{\mu}$, thus
$$
DU^{\mu}=U^{\mu}\left( x^{\alpha}+dx^{\alpha}\right) -U^{\mu\prime}=dU^{\mu
}-\delta U^{\mu},
$$
where $dU^{\mu}=U^{\mu}\left( x^{\alpha}+dx^{\alpha}\right) -U^{\mu}\left(
x^{\alpha}\right) $ is the ordinary differential. Therefore, the small addition
$\delta U^{\mu}$ is the result of parallel translation. There are two obvious
properties of $\delta U^{\mu}$: it should be linear in $U^{\mu}$ and should
vanish with $dx^{\mu}\rightarrow0$. Therefore one can represent $\delta
U^{\mu}$ as follows:
$$
\delta U^{\mu}=-\Gamma_{\alpha\beta}^{\mu}U^{\alpha}dx^{\beta},
$$
where $\Gamma_{\alpha\beta}^{\mu}$ is the set of some matrices usually referred
as “connection coefficients” or “Christoffel symbols”.
Using $\Gamma$, one can generalize the covariant differential $D$ to any tensor quantities.
Although, there are no additional mathematical requirements on $\Gamma$, there
are physical ones in GR — the equivalence principle requires that $\Gamma$
should be symmetric $\Gamma_{\alpha\beta}^{\mu}=\Gamma_{\beta\alpha}^{\mu}$
and $Dg_{\alpha\beta}=0$. The last condition results in
$$
\partial_{\mu}g_{\alpha\beta}=-\left( \Gamma_{\mu,\beta\alpha}+\Gamma
_{\beta,\mu\alpha}\right) ,
$$
where $\Gamma_{\mu,\beta\alpha}=g_{\mu\rho}\Gamma_{\beta\alpha}^{\rho}$. Using
the condition that $\Gamma$ is symmetric one can find:
$$
\Gamma_{\beta\alpha}^{\rho}=\frac{1}{2}g^{\rho\sigma}\left( \partial_{\beta
}g_{\sigma\alpha}+\partial_{\alpha}g_{\sigma\beta}-\partial_{\sigma}
g_{\alpha\beta}\right).
$$
Let's now consider a parameterized trajectory $x^{\mu}\left( \lambda\right)$, the contravariant vector called speed is $U^{\mu}=dx^{\mu}\left(\lambda\right)/d\lambda$, therefore the contravariant acceleration takes the
form:
$$
A^{\mu}=\frac{DU^{\mu}}{d\lambda}=\frac{dU^{\mu}}{d\lambda}-\frac{\delta
U^{\mu}}{d\lambda}=\frac{dU^{\mu}}{d\lambda}+\Gamma_{\alpha\beta}^{\mu
}U^{\alpha}U^{\beta}.
$$
If we choose $d\lambda=d\tau$ then in the locally-inertal frame ($\Gamma=0$)
for the trajectory $x^{\mu}\left( \lambda\right) $ the acceleration $A^{\mu
}$ coincides with the ordinary one $\left( 0,\mathbf{a}\right) $.
And vice versa, $DU^{\mu}=0$ means that $U^{\mu}$ is constant in the
locally-internal frame although it does imply that it is constant in any
other frame, in fact $dU^{\mu}=\delta U^{\mu}$ implies that a free-fall
trajectory is actually a parallel translation in GR, which (for an external
observer) looks like the action of gravitational forces.
