The "slope of the secant line" interpretation makes perfect sense for continuous functions and anyone familiar with the derivative knows its applications. For example, if $f(x)=x^2$, then its average rate of change (AROC) over the interval $[1, 2]$ is $f_{\rm AROC}=\frac{f(2)-f(1)}{2-1}=3$, and this can be used to roughly approximate the derivative of $f$.
But what about a discontinuous function, such as $g(x)=\begin{cases}0&\text{for }x<\frac12\\1&\text{for }x\ge\frac12\end{cases}$ over $[0, 1]$, or $h(x)=\frac1x$ over $[-1, 1]$? Does the notion of AROC make sense, so that $g_{\rm AROC}=\frac{1-0}{1-0}=1$ and $h_{\rm AROC}=\frac{1-(-1)}{1-(-1)}=1$?
I cannot seem to find a formal definition for the AROC, although some textbooks and calculators like Wolfram|Alpha express the AROC of $f(x)$ in terms of the average value of $f'(x)$, e.g.
$$f(x)=x^2,x\in[1,2]\\ \implies f_{\rm AROC}=\frac1{2-1}\int_1^2\frac{\mathrm d(x^2)}{\mathrm dx}\,\mathrm dx=2\int_1^2x\,\mathrm dx=3$$
and this discussion usually accompanies the mean value theorem, which does not apply to discontinuous functions.
The best I can come up with is defining $\lambda(x,y)=\frac{f(y)-f(x)}{y-x}$ and averaging it, using $$f_{AROC}=\frac{1}{\#U}\iint_U\lambda(x,y) dy dx$$
where I use $\#U$ to represent the length, area, volume, size etc of $U$.
For example, when $f: [1,2]\to [2,4] ; f(x)=x^2$, then $\lambda : [1,2]\times [1,2] \mapsto [2,4]; \lambda(x,y)=x+y$, and $$f_{AROC}=\frac{1}{1\cdot 1}\int_1^2\int_1^2 (x+y) dy dx =3$$
Applying this to $g$, we observe that: $$\lambda_g(x,y)=\begin{cases}\frac{-1}{y-x} & x\geq \frac 12, y< \frac 12 \\ \frac{1}{y-x} & x<\frac12, y\geq \frac12 \\ 0& \text{otherwise} \end{cases}$$
Again the area is $1\cdot 1=1$, and $$g_{AROC}=\bigg[\int_0^\frac 12 \int_\frac 12^1 \frac{1}{y-x}dy dx+\int_\frac12^1 \int_0^\frac 12 \frac{-1}{y-x} dy dx\bigg]=2\ln2$$
As seen here, note both double integrals are equal (but nasty!)
As for $h$, that integral does not converge, as is to be expected, due to the graph's asymptote at $x=0$