= Solution
In general coordinates the basis changes along the <world line>. Differentiating only the components of a <vector> omits that basis change, and does not transform as a <vector> under a position-dependent coordinate transformation. The appropriate rate is the <covariant derivative along a curve>:
$$
\boxed{\frac{DV^\alpha}{d\lambda}
=\frac{dV^\alpha}{d\lambda}
+\Gamma^\alpha{}_{\beta\gamma}\frac{dx^\beta}{d\lambda}V^\gamma.}
$$
For the metric-compatible torsion-free connection, the <Christoffel symbols> are determined by the <metric tensor> and its first derivatives:
$$
\Gamma^\alpha{}_{\beta\gamma}
=\frac12g^{\alpha\delta}(\partial_\beta g_{\delta\gamma}
+\partial_\gamma g_{\delta\beta}-\partial_\delta g_{\beta\gamma}).
$$
Thus the ship's <four-acceleration> is
$$
\boxed{a^\alpha=\frac{Du^\alpha}{d\tau}
=\frac{du^\alpha}{d\tau}+\Gamma^\alpha{}_{\beta\gamma}u^\beta u^\gamma.}
$$
In Cartesian coordinates of an <inertial frame> in flat <spacetime>, the connection coefficients vanish and the ordinary component derivative happens to suffice. In curved <spacetime> it suffices at a chosen event in a <local inertial frame>, but not generally along an extended trajectory.
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