The instantaneous displacement is just the displacement of the system evaluated at some instant $t$, $x(t)$.
As we intuitively think the speed of a body in rest should be zero, it is not so intuitive to think the displacement of a body from a reference point should be zero too. Since our intuitive reference for speed is zero, we should also expect our intuitive reference for displacement to be zero too.
If you think in the mass+spring+damper system: $$ \ddot x+2\zeta \omega\dot x+\omega^2x=0, x(0)=1 $$ Its zero energy state is at $x=0$ and $\dot x=0$, when both the instantaneous velocity and the instantaneous displacement are zero. In this case, the intuition is somehow more explicit.
In your specific example, if $x(t)=\frac{t^2}2+t$, the instantaneous displacement at $t=2$ is $x(2)=4$, meaning the instantaneous displacement measured from the reference is 4.
In the case of the mass spring damper, for example, if we have an underdamped system $x(t)=e^{-\zeta\omega t}\cos(\omega\sqrt{1-\zeta^2}t)$, the same concept applies, the instantaneous displacement at $t=0$ is $x(0)=1$ but in here it is more evident that the displacement is a deviation from the reference.
In contrast, check the following: Suppose we are measuring a displacement with respect to the last measured position, i.e. suppose $dx'(t)=dt'$ represents the change of displacement from the last time with a resolution of $dt'$, that is, after each $dt'$ interval, the displacement increases in that same measure. In this case, we should, again, integrate to obtain the displacement: $$ x(t)=\int_0^{t} dx'=\int_0^t dt'=t $$ That is, since the position is constantly changing, we expect the position to be linearly increasing. Again, $x(t)$ represent a measurement wrt to a reference, and $\delta x(t)$ is a mathematical object to depict what you could understand by change of displacement. Note, how this last concept is by far less used, hence the use of that notation, less used too.