I'm going to turn my comment into an answer.
The error order tells you up to which order (exclusive) the discrete
solution corresponds to the exact solution.
An $\mathcal{O}(\Delta t^4)$ method reproduces the orders 0-3 of the Taylor expansion of the
function around $\Delta t$ (i.e. in the expression $f(t+\Delta t) = f(t) +\dots$).
Put bluntly (and mathematically imprecisely)
if you halve the size of the time step $\Delta t$ the deviation will be smaller by $2^4$.
This, of course, assumes that the error is computed at the same final time.
Or, if you want to use the log-log plot effectively you can say that by reducing
the step size by one order of magnitude $\Delta t = 10^{-1} \rightarrow \Delta t = 10^{-2}$
the deviation from the analytic solution will be reduced by 4 orders of magnitude, i.e.,
instead of $10^{-2}$ it will be $10^{-6}$.
To visualize the growth of an error one could plot curves of different orders ($\Delta t^2, \Delta t^3$ etc.)
and try and compare the result to these curves. This is very finnicky in a log-log plot and
results in a rather confusing plot.
A better way is to define a function that will draw a slope triangle with appropriately
labelled sides close to the curve(s) you want to show the behaviour of.
I have used this in the analysis of the implementation of $\mathcal{P}_2,\mathcal{P}_3$
finite elements in my thesis, as can be seen in the following plot.

Note that this plot would be really confusing if I were to add an entire curve for each growth order. Admittedly, the slope triangles could have been placed better, but they do show the growth order (one power of $2$ of mesh width corresponds to $n$ powers of $10$ on the error axis).