The displacement is only the velocity multiplied by the elapsed time if velocity is constant as you suggested. To derive the equation for varying velocity you must consider the infinitesimal case where the elapsed time is so small that you can consider velocity constant. In this case, a small displacement $dx$ is given by the product of velocity v(t) by small increment of time $dt$. So, the correct derivation would be:
\begin{equation}
dx = v \, dt
\end{equation}
\begin{equation}
\int_{x_0}^{x(t)} dx = \Delta x = x(t) - x_0 = \int_{t_0}^{t_F} v(t) \, dt = \int_{t_0}^{t_F} v_0 + a\, t \, dt
\end{equation}
\begin{equation}
x(t) = x_0 + v_ 0 \, \Delta t + a \frac{\Delta t^2}{2}
\end{equation}
with $\Delta t$ = $t_F$ - $t_0$.
You can visualize this graphically by thinking on the velocity by time graph. As velocity increases linearly with time, the area below the graph will be a triangle with height $a\, \Delta t$ and width $\Delta t$ plus a rectangle with height $v_0$ and width $\Delta t$. So the area is:
\begin{equation}
Area = \Delta x = x(t) - x_0 = v_0 \, \Delta t + \frac{(a \, \Delta t) \, \Delta t}{2}
\end{equation}