This would not be as confusing if one were to do it the "mathematics way", where there is no abuse of notation like $V^\mu$ ($V^\mu\partial_\mu$ is a vector but $V^\mu$ is the component of the vector), but let's stick to the common GR convention.
The distinguishing feature is that although $\mu=0,1,2,3$ in both cases, $V^\mu$ refers to a (tangent) vector while $x^\mu$ refers to coordinates of a spacetime event. More formally, let $(M,g)$ be the spacetime with metric $g$ and $p\in M$ an event in spacetime. On one hand, $x^\mu(p)$ gives the coordinates of that point, and more generally given a curve $\gamma: \mathbb{R}\to M$, we have that $x^\mu(\gamma(\tau))$ are spacetime coordinates along the curve. This is often made into a shorthand $x^\mu(\tau)$ where the curve is clear from the context.
On the other hand, the four-velocity is a tangent vector in the sense that for the same curve,
\begin{align}
\dot{\gamma}(\tau) = \frac{d\gamma(\tau)}{d\tau} \in T_{\gamma(\tau)}M\,.
\end{align}
In local coordinates, by chain rule we have roughly speaking (I am slightly sloppy for now)
\begin{align}
\dot{\gamma}(\tau) \equiv \frac{d x^\mu}{d\tau}\frac{\partial }{\partial x^\mu}
\end{align}
so the four-velocity components are $U^\mu=dx^\mu/d\tau$.
Operationally, this is what happens: covariant derivatives acting on functions reduces to partial derivatives, while acting on vectors it does not. Coordinates $x^\mu$ are functions on the manifold since for each $\mu$ we have $x^\mu: M\to \mathbb{R}$. Therefore, by definition we have
\begin{align}
\nabla_\mu x^\nu = \partial_\mu x^\nu = \delta_\mu^\nu\,,
\end{align}
but $U^\mu$, being (in GR convention) a vector, is treated differently, namely
\begin{align}
\nabla_\mu U^\nu = \partial_\mu U^\nu + \Gamma^\nu_{\mu\sigma}U^\sigma\,.
\end{align}
Now to answer your question. Essentially, in practice this means that if $V^\nu$ refers to a vector, then
\begin{align}
\frac{d}{d\tau}V^\nu &\equiv \frac{dx^\mu}{d\tau}\frac{\partial V^\nu}{\partial x^\mu} = U^\mu\partial_\mu V^\nu\,,\\
\frac{D}{D\tau}V^\nu &:= U^\mu\nabla_\nu V^\nu =U^\mu(\partial_\mu V^\nu + \Gamma_{\mu\sigma}^\nu V^\sigma) \,.
\end{align}
Note that this means for coordinate functions we really have
\begin{align}
\frac{D x^\mu}{D\tau} = \frac{d x^\mu}{d\tau}\,,
\end{align}
while for vectors we have
\begin{align}
\frac{D V^\nu}{D\tau} = \frac{d V^\nu}{d\tau}+U^\mu \Gamma^{\nu}_{\mu\sigma}V^\sigma\neq\frac{d V^\nu}{d\tau} \,.
\end{align}
Therefore, in geodesic equation the second derivative is really $d^2/d\tau^2$ because the Christoffel symbols have been separated out:
\begin{align}
0 = \frac{D^2 x^\mu}{D\tau^2} = \frac{D}{D\tau}\frac{d x^\mu}{d\tau} = \frac{D U^\mu}{D\tau} = \frac{dU^\mu}{d\tau}+\Gamma^\mu_{\nu\sigma}U^\nu U^\sigma\,.
\end{align}
As for your second question, the requirement of $U^\mu U_\mu = -c^2$ must hold for any timelike 4-velocity of any observer. A worldline of an observer is a geodesic if and only if it is "free-falling", in the sense that the observer is not acted upon by external force. For example, in presence of electromagnetic field, a charged particle will not move along geodesics as the EM field will exert a four-force on the particle. In this sense the charged particle is not moving inertially.
