The most intuitive way of understanding acceleration is to understand it in terms of Taylor series expansions
$$\sum_{n=0}^{\infty} \dfrac{f^{(n)}(u)}{n!} (t-u)^n$$
A good entry level discussion on how one applies the Taylor series expansion to the question of position-velocity-acceleration can be found in this short paper from this website attributed to S.A. Fulling.
If we review Taylor's Theorem we start out with evaluating the series with $u=0$, Wolfram shows the expansion as:
$$f(t) = f(0) + tf'(0) + \dfrac{t^2}{2!}f''(0) + \dots + \dfrac{t^{(n-1)}}{(n-1)!}f^{(n-1)}(0) + \int_0^t \dfrac{(t-u)^{(n-1)}}{(n-1)!}f^{(n-1)}(u) du$$
If you look at the first three terms, you should see the similarity to:
$$x(t) = f(t) =x_0 + v_ot + \dfrac{1}{2}a_0t^2$$
Now first consider the expansion of the $\frac{1}{n!}$ inverse factorial terms.
$$\dfrac{1}{0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \dfrac{1}{5!} + \dots = \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{24} + \dfrac{1}{120} + \dots$$
then consider the sum of only the terms after the $\dfrac{1}{2}$ term:
$$\sum_{n=3}^{\infty} \dfrac{1}{n!} = \lim_{n \to \infty} \dfrac{1}{n!} = e - 2.5 = 0.218282\dots$$
The summation of all the inverse factorial terms after $\frac{1}{2!}$ is only $0.218282\dots$ which is clearly less than $\frac{1}{2}$, so unless the higher order derivatives of $f^{n}(u)$ when $u=0$ are substantial, then the overall effect of the higher order derivatives will never be greater than the first non-linear factor $t^2$.
Returning to the Taylor expansion, the goal of course is determining the values of $x_0$,$v_0$,$a_0$. These are frequently given, or can be determinable by observation, but in any case can be understood as constants obtained from integration. For instance, as explained in the above exercise, one would start by integrating some arbitrary constant for the third derivative of some function assuming the real integral was less than the integral of an arbitrary constant:
$$\int_0^t f'''(u) du \le \int_0^t M du$$
such that
$$f''(t) + a_0 \le Mt$$
performing these integrals for successive derivatives we can eventually find the form of the equation,
$$|f(t)-x_0 + v_0t + \dfrac{1}{2}a_0t^2| \le \dfrac{1}{6}M|t|^3$$
Per the exercise, this shows that the the graph of $f(t)-x_0 + v_0t + \frac{1}{2}a_0t^2$ will be between two curves $\pm \frac{1}{6}M|t|^3$ which should be very close together at $t=0$.
Since $M = f'''(0)$, if $$f'''(0) = \epsilon$$ with $$\epsilon \approx \frac{1}{\infty}$$ then we could say
$$f(t)\approx x_0 + v_0t + \frac{1}{2}a_0t^2$$
In addition, if the derivative expansion is shown to have higher order terms that cancel, or are sufficiently suppressed, then the higher order terms can be ignored.
However, the specific values of the $x_0$,$v_0$,$a_0$ will not be known unless domains of integration and boundary conditions are specified.
In any case, one can continue to expand the position variable with respect to time and find that "jerk" is the third power or time and "Jounce" (or "Snap") is the fourth power in the expansion. These higher order contributions, and all other higher order contributions, have a decreasing contribution to the overall equation attributable to the $n!$ factorial term in the denominator of the summation. This is a good analogy to understanding the concept of the coupling constant and how the coupling can decrease as one expands perturbatively about a solution to an equation (e.g. a function).
So acceleration is the first non-linear term in the expansion of the function relating position to time. Since higher order terms do not contribute more than the first non-linear term (acceleration) for certain $f(t)$, their effects can be ignored in most circumstances (although in some situations, such as elevator design, jerk is a consideration, and when one gets into design space craft, the higher order derivatives must be considered as well).
It should be noted that $f(t)$ has units of position (e.g. m = meters), since the Taylor expansion is for the $t$ variable, and the $t$ variable is raised to a higher order in each successive term, the constant must carry units which cancel out the units of time for each term. Since $a_0$ is associated with $t^2$, it must carry units of $\frac{m}{s^2}$ in order to ensure that $f(t)$ is in units of $m$.