$\vec{s}$ is an arbitrary position vector, which can change with time.
$\vec{v} = \dfrac{\Delta \vec{s}}{\Delta t}$ is the velocity vector.
$\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t}$ is the acceleration vector, and it readily follows that \begin{align*} \vec{v_f} & =\vec{v_i}+\vec{a}\Delta t\\ \dfrac{\Delta \vec{s}}{\Delta t} & =\vec{v_i}+\vec{a}\Delta t\\ \Delta \vec{s} & =\vec{v_i}\Delta t+\vec{a}(\Delta t)^2 \end{align*}
However, this differs from the accepted $\Delta x=v_0t+\frac{1}{2}at^2$.
What is wrong with my reasoning, and how can I get to the correct answer using purely vectors?
Your second to last step is the statement $$\frac{\Delta{\bf s}}{\Delta t}={\bf v}_f$$ which doesn't make a lot of sense. One could argue that it makes more sense to use the velocity at the middle of the time interval $\Delta t$, in which case you get the desired result.