Simple functions assume finitely many values in their image, and can be written as
$$f(\omega)= \sum_{i=1}^n a_i \mathbb I_{A_i}(\omega), \quad \forall \omega \in \Omega$$
where $a_i \geq 0, \forall i \in \{1,2,3, \dots, n\},$ and $A_i \in \mathcal F, \forall i.$
So this is how I process it in "human": For each outcome in the sample space (i.e. $\omega$), one must check whether it belongs or not to a measurable set $A_i$ in the sigma algebra $\mathcal F.$ If the Boolean operation (characteristic function $\mathbb I_{A_i}$) is $1,$ the result of the function will be some value $a_i$ which will be exactly the same for all outcomes in $A_i.$ This could be symbolically plotted as step function, which each step corresponding to one of the combinations of the subset $A_i$ and the value $a_i$ of all outcomes $\omega \in A_i$.
So far, clear as day.
Now, when you introduce the $\sum$ at the beginning of the definition, it looks like you are integrating: in other words, the function $f(\omega)$ with the sum in front doesn't seem to "spit out" the corresponding step of that particular $\omega,$ but rather all the steps for all omegas - all at once. And that "all at once" seems like a contradiction: after all in a "truly simple function," such as $f(x)=2x+2,$ you don't get a line because you sum the results of the function across the real line, but because you collect as a set the results for every and each value of the real line entered into the function as an independent variable forming a line.
The answer is that given the finite values in the image of the simple function $f,$ corresponding to each of $n$ disjoint subsets of $\Omega,$ such that $\bigcup_{i}A_i=\Omega,$ it can be written as:
$$f(\omega)=a_1 \mathbb I_{A_i}(\omega) + a_2 \mathbb I_{A_2}(\omega)+\cdots +a_n\mathbb I_{A_n}(\omega) =\sum_{i=1}^n a_i \mathbb I_{A_i}(\omega).$$
So it is a nice trick to compactly notate that the only $a_i$ that should survive in the sum is the one that matches the subset $A_i$ according to the $i$ indexation, akin to a for-loop call in programming: only the $a_i$ corresponding to $\mathbb I_{A_i} (\omega)=1,$ i.e. when the outcome is in the measurable set $A_i,$ is ultimately represented. Without the sum this would have to be defined differently, or left implied.