**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.

The 4-velocity $\mathbf u$ of a particle with mass is always normalised so that
$g_{\mu\nu}u^\mu u^\nu=-1$ (I am using a -+++ metric signature, you want $+1$ otherwise). From this you can express the $u^0$-component in terms of the other components, up to sign. Then pick the option which has $\mathbf u$ pointing forward in time.

## Best Answer

Intuitively, geodesics are paths that objects under only the action of the gravitational field follow. The derivative of the four velocity being zero roughly means there is "no acceleration" in the covariant sense; ie no external forces. These objects are essentially just rolling around in spacetime, moving in the direction specified by the curvature.

There is a lot more to your question; for instance, you probably mean something more like the following local expression

$$\mathcal{D}_\nu U^\mu=\partial_\nu U^\mu+\Gamma^{\mu}_{\rho\nu}U^\rho=0$$

where I am using the Christoffel symbols to explicitly write the connection in the tangent bundle (notice this is now a $(1,1)$-tensor). This is what we mean by "directional derivative", since this is now saying that "the velocity does not change in the direction $\nu$."