Although the pure mathematical concept of vector is more abstract than this, in physics, especially Newtonian, by vectors we usually mean equivalence classes of directed line segments under the equipollence relation, which is Bellavitis' definition.
How does the zero vector fit this definition? It's not a directed line segment, since line segments have by definition non-zero length. Even if we extend the definition to be equivalence classes of directed line segments and directed points, instead of just the former, not all problems are solved - all such points would have to be equipollent, since there is only one zero vector. By definition they have the same length, but I don't see how they have equal direction - zero vector has "arbitrary direction", which I assume, correct me if I'm wrong, that one can arbitrarily assign its representative any direction. But if this is correct, then such points would have different directions, and wouldn't form an equivalence class under equipollence, leading to a separate zero vector for each direction. We know, however, that there is only one zero vector. In what way are these problems solved? How does zero vector fit Bellavitis' definition? If it doesn't, how could the definition be extended?
The simplest way to handle this is to change your definition of equipollence to say that two directed segments are equipollent if one can be obtained by a translation of the other. In particular, this definition does not actually involve any reference to "direction", but is equivalent to "same direction and length" if your segments are nondegenerate. There is now no problem with allowing your "line segments" to be degenerate (i.e., the endpoints are the same). Any two degenerate line segments are indeed equipollent by this definition, since you can just take the translation which sends one point to the other point, so there is a unique vector of length $0$.