What is displacement? Position relative to a reference point or change of position

It's possible that the way in which these terms are used varies from person to person, even among professionals in the field. However, in the usage I'm familiar with, displacement is the change in position, period. Definition #2 is correct and #1 is wrong. (The length and direction of a line from a fixed reference point is just called position.)

In this usage the proper form of the constant acceleration kinematic equation would be $\Delta \vec{x} = \vec{v}t + \frac{1}{2}\vec{a}t^2$, or $\vec{x} = \vec{x}_0 + \vec{v}t + \frac{1}{2}\vec{a}t^2$, where $\vec{x}$ is position and $\vec{v}$ is initial velocity. It would be valid to write $\vec{x} = \vec{v}t + \frac{1}{2}\vec{a}t^2$ if you always choose the origin to be at the initial position, but that seems like an unnecessary restriction.

Alternatively, you could write the equation in terms of displacement. If you use $\vec{s}$ for displacement, the equation would be $\vec{s} = \vec{v}t + \frac{1}{2}\vec{a}t^2$. That is because $\vec{s} = \Delta\vec{x}$ (displacement equals change in position). If these textbooks you're using are using this notation in which $\vec{s}$ is displacement, then it seems very strange to write $\vec{v} = \Delta\vec{s}/\Delta t$. That is unconventional and probably unclear notation, although it might not necessarily be wrong.

The Wikipedia usage is fine, though, because in that formula the displacement is $\Delta\vec{d}$, not just $\vec{d}$. In $\Delta\vec{d} = \vec{d}_f - \vec{d}_i$, the vectors $\vec{d}_i$ and $\vec{d}_f$ could be either positions or displacements.