Why do we do partial and not covariant differentiation with $x^{\nu}$? Why when taking the velocity vector we make
$$u^{\nu}=\frac{d}{d\tau}x^{\nu}$$
and not
$$u^{\nu}=\frac{\nabla}{d\tau}x^{\nu}$$
where in the last equation I meant the covariant derivative. Why?
 A: Consider an $n$-manifold $M$. A curve is simply a continuous map $\gamma : \mathbb{R} \to M$. For simplicity, suppose $M$ is covered by a single coordinate chart (diffeomorphism) $\varphi : M \to \mathbb{R}^n$. Putting these together, we have
\begin{eqnarray}
\mathbb{R} & \stackrel{\gamma}{\longrightarrow} & M & \stackrel{\varphi}{\longrightarrow} & \mathbb{R}^n \\
\tau & \longmapsto & p & \longmapsto & (x^0, x^1, \ldots, x^{n-1})
\end{eqnarray}
Thus we can view $x^0$ as a simple map $\mathbb{R} \to \mathbb{R}$, and likewise for $x^1$, etc. Each such function is differentiated in the normal way (partial or total differentiation, the two being the same for functions from $\mathbb{R}$ to $\mathbb{R}$), and really there is no other derivative to define.
The concatenation $\varphi \circ \gamma$ circumvents the manifold entirely, relying only on it having some minimal topological structure, not differential (metric) structure. Another way of looking at the situation is to think of $\gamma(\mathbb{R})$ as a one-dimensional submanifold of $M$. As a 1D manifold, it doesn't have intrinsic to itself all the curvature it gets as being embedded in $M$. In fact, as a 1D manifold, it has no intrinsic curvature.
The covariant derivative is only defined for tensors. Let's focus on a fixed $p$ in $M$, for which there is a tangent space $T_p(M)$. Assuming we've defined coordinates, the natural choice of basis for this $n$-dimensional vector space is the set of partial derivatives with respect to those coordinates: $\{\partial/\partial x^0, \partial/\partial x^1, \ldots, \partial/\partial x^{n-1}\}$. And of course tensors are just linear maps from products of $T_p(M)$ and its dual into $\mathbb{R}$.
If you want to covariantly differentiate the object with components $x^\mu$, then you should be able to write it as a linear combination of $\{\partial/\partial x^0, \partial/\partial x^1, \ldots, \partial/\partial x^{n-1}\}$. But this can't be done sensibly -- coordinates and the directional derivatives they induce are entirely different beasts.
