I should state that the following steps depend on the formula for a geometric series:
$$\sum_{j=1}^{k} r^{j-1} = \frac{1-r^k}{1-r}$$
The sum in question is
$$p_1 p_2 \sum_{j=1}^{k-1} (1-p_2)^{j-1} \sum_{i=1}^{k-j} (1-p_1)^{i-1} $$
Evaluate the inner sum first:
$$\sum_{i=1}^{k-j} (1-p_1)^{i-1} = \frac{ 1 - (1-p_1)^{k-j} }{p_1} $$
Now the sum is
$$\begin{align} p_2 \sum_{j=1}^{k-1} (1-p_2)^{j-1} \left [ 1 - (1-p_1)^{k-j} \right ] &= 1 - (1-p_2)^{k-1} - p_2 \sum_{j=1}^{k-1} (1-p_2)^{j-1} (1-p_1)^{k-j} \\ \end{align} $$
The sum on the right-hand side may be evaluated as follows:
$$\begin{align}\sum_{j=1}^{k-1} (1-p_2)^{j-1} (1-p_1)^{k-j} &= (1-p_1)^{k-1} \sum_{j=1}^{k-1} \left ( \frac{1-p_2}{1-p_1} \right )^{j-1} \\ &= (1-p_1)^{k-1} \frac{ 1 - \left ( \frac{1-p_2}{1-p_1} \right )^{k-1}}{1 - \frac{1-p_2}{1-p_1}} \\ &= (1-p_1)\frac{(1-p_1)^{k-1} - (1-p_2)^{k-1}}{p_2-p_1} \\ \end{align} $$
Now we can put this all together:
$$\begin{align} 1 - (1-p_2)^{k-1} - p_2 (1-p_1)\frac{(1-p_1)^{k-1} - (1-p_2)^{k-1}}{p_2-p_1} \\ = 1 - \frac{(p_2-p_1)(1-p_2)^{k-1} + p_2 (1-p_1)[(1-p_1)^{k-1} - (1-p_2)^{k-1}]}{p_2-p_1} \\ \end{align}$$
which, after some cancellation and consolidation, produces the following result for the sum:
$$1 - \frac{p_1 (1-p_2)^k - p_2 (1-p_1)^k}{p_1-p_2}$$
and is equal to the result stated above.
The work with Iverson brackets looks (besides notational aspects) fine. Note that in the derivation we not only have to show implications $\rightarrow$ but equivalence relations of the index regions we want to sum up.
Here is a derivation based upon index transformations without using Iverson brackets.
We obtain
\begin{align*}
\color{blue}{\sum_{n=0}^{\infty}\sum_{m=0}^n\xi(m,n)}
&=\sum_{0\leq m\leq n<\infty}\xi(m,n)\tag{1}\\
&=\sum_{{0\leq m}\atop{2m\leq m+n<\infty}}\xi(m,n)\tag{2}\\
&=\sum_{{0\leq m}\atop{2m\leq l<\infty}}\xi(m,l-m)\tag{3}\\
&=\sum_{0\leq l<\infty}\sum_{0\leq 2m\leq l}\xi(m,l-m)\tag{4}\\
&\,\,\color{blue}{=\sum_{l=0}^\infty\sum_{m=0}^{\lfloor l/2\rfloor}\xi(m,l-m)}
\end{align*}
and the claim follows.
Comment:
In (1) we write the index region somewhat more conveniently.
In (2) we do an equivalence transformation of the inequality chain by adding $m$ as preparation for the next step.
In (3) we introduce $l$ and substitute $l=m+n$.
In (4) we rewrite the index region, so that summing over $l$ becomes the outer series.
Best Answer
I would evaluate this expression in the following way (in order):
$$\tag{1} \sum_{y=1}^{M} \left(1 + a\cdot f\right)(1 + b \cdot f\left(y\right)) $$
I'm certain that there's no way to "simplify" your expression without knowing the functional form of $M(x)$, for example is it $M(x)=\cos(x)$ or is it $M(x)=x^2 - \log(x) + \max(x,y,f(x),f(y),f(x)-f(y))$?