In my introductory differential geometry course (in which we only really dealt with $\Bbb{R}^3$ apart from some 'asides' to give context and to give some intuition about generalising some concepts for the future), we briefly covered geodesics in the last week of term. The definition given was as follows:
Let $M \subset \Bbb{R}^3$ be a surface, with $U$ a unit normal vector field on $M$,
$\alpha \colon I \to M$, such that $\alpha$ is smooth, is a geodesic if:
$\forall t \in I$, $\alpha''(t)$ is parallel to $U(\alpha(t))$.
Given this definition, it strikes me that $\vec{0}$ should be considered a geodesic, since $\vec{0}$ is considered parallel to every vector, and as such, setting $\alpha = \vec{0}$ gives us that $\alpha''(t) = \vec{0}$ and thus $\vec{0} \parallel U(\alpha(t)) = \vec{0}$.
It also strikes me that, given a geodesic is also described informally in my course as the shortest path between two points on a surface in $\Bbb{R}^3$, it is somewhat necessary to consider the zero vector as a geodesic, as it is of course always the shortest path between any point and the point itself
Is this true? If not, where have I gone astray in my reasoning? Furthermore, does this generalise to every geometric space? (Intuitively, it seems to me that it should)
As your question is written you're calling a single vector a path, and those are two different things.
It would be correct if you defined a path $\alpha(t) = x_0 \forall t$, i.e. a constant path. Any such $\alpha$ would be a geodesic. Note that it is possible for some definitions of geodesic to explicitly rule out such trivial paths, but without such a limitation it would be.