The "average" in average rate of change comes from sum of instantaneous rate of change divided by number of rates of change? To show what I mean here is a graph of $y = x^2$.

The red line represents AROC from $a$ to $b$.
The blue lines represent the IROC at some points $x$, where $a<x<b$
If I were to calculate the gradients of those blue lines, add them up and divide by the number of lines I used, would I get the average rate of change? If this is true; is this why the "average" exists in average rate of change?
 A: The answer is a qualified "yes".
I say "qualified" because you have to be careful that you add up the lines in a way that fairly reflects the way you drew them.
Also, to be sure of getting an answer that exactly agrees with the "average",
rather than only approximately agreeing, you may have to take a limit where you reduce the space between $x$ points toward zero.
For an example of being careful about how you add up the lines, if you took $a=-1$ and $b=1,$ the average rate of change is zero.
If you take gradients of blue lines at points chosen symmetrically around the $y$ axis you will get the same result,
but if you take, say, gradients at $x=-1$ and $x=-\frac12$ and then take gradients at a hundred positive values of $x$ at intervals of $0.01$ between $x$ values,
you will get a lopsided result.
You can guard against such a lopsided result in either of two ways: you can insist that the gradients be taken at points uniformly spaced between $a$ and $b,$
or you can take a weighted average where the weight of each gradient depends on how far away the nearby gradients were taken.
Regarding the possible need to take a limit,
your observation is essentially the Fundamental Theorem of Calculus.
That is, if you are looking at a graph of $y=F(x)$ where $F$ is a differentiable function,
then the gradient at each point of the graph is $F'(x).$
Define the function $f$ by $f(x) = F'(x).$
Then
$$ F(b) - F(a) = \int_a^b f(x)\,\mathrm dx. $$
Now suppose you evaluate the integral using a Riemann sum with uniform intervals.
The sum with $n$ intervals is the sum of $n$ terms
of the form $f(x_i) \Delta x,$ where $\Delta x = (b - a)/n$:
$$ f(x_1)\frac{b-a}n + f(x_2)\frac{b-a}n + \cdots + f(x_n)\frac{b-a}n 
 = \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n} (b - a). $$
That is, the integral is just $b - a$ times the average of the instantaneous $f(x_i)$ values,
whereas $F(b) - F(a)$ is just $b - a$ times the average rate of change.
But as you may be aware, you don't always get the exact value of an integral on the first try when doing a particular Riemann sum with some finite number of terms.
You may need to look at what happens in the limit as $n \to \infty$
in order to get an accurate value of the integral.
A: Average rate of change of a function $f$ over an interval $[a,b]$ is defined to be $$\frac{f(b)-f(a)}{b-a}$$ This gives the slope of the line joining $(a,f(a))$ and $(b,f(b))$. It signifies what steady (constant) rate of change of the function would be required to get from $(a,f(a))$ to $(b,f(b))$
For instance if $f(3)=11$ and $f(5)=23$, then over the length $2$ interval from $x=3$ to $x=5$, the function changed by $23-11=12$ units. This amounts to an average of $\frac{12}{2}=6$ units of change in $f$ per unit of change in $x$ over the given $x$-interval.
