Here is an interesting book. If the download doesn't start copy and paste that link to your browser. Indeed, I'm unable to apply the Inclusion-Exclusion principle to an instance of it. How have we obtained the lower bound here and how would it continue with more terms ?
$$\sum_{y\in R(x)} \text{Pr}_{h\in H^{m_x}_l}[h(y)=0^{m_x}]-\sum_{y_1<y_2\in R(x)} \text{Pr}_{h\in H^{m_x}_l}[h(y_1)=h(y_2)=0^{m_x}]$$.
My confusion lies in the fact that I would thing that we should extract from first term yet this and so that the result would look like this
$$\sum_{y\in R(x)} \text{Pr}_{h\in H^{m_x}_l}[h(y)=0^{m_x}]-\sum_{y_1<y_2\in R(x)} \text{Pr}_{h\in H^{m_x}_l}[h(y_1)=h(y_2)=0^{m_x}]-\color{red}{\sum_{y_1<y_2\in R(x)} \text{Pr}_{h\in H^{m_x}_l}[h(y_1)=h(y_2)=1^{m_x}]}$$
You are correct in that each term in the summation of the inclusion-exclusion principle "extracts" (by which I assume you meant subtracts) from the previous term, but when you distribute the negations in the sum certain terms end up being positive! Let's say I have some events $E_1, E_2,...,E_k$, and I wish to lower bound $\Pr(\bigcup^k_{i=1}E_i)$. The Inclusion-Exclusion principle tells us:
$$\Pr(\bigcup^k_{i=1}E_i) = \sum_{1 \leq i_1 \leq k}\Pr(E_{i_1}) \color{red}{-} \left( \sum_{1 \leq i_1 < i_2 \leq k}\Pr(E_{i_1} \cap E_{i_2}) - \color{blue}{\left( ... \sum_{1 \leq i_1 < ... <i_{l} \leq k}\Pr(\bigcap^{l}_{j=1} E_{i_j}) - \left(...\right) ...\right)} \right)$$
for $l \in [k]$. However, if my goal is simply to lower bound $\Pr(\bigcup^k_{i=1}E_i)$, then I can stop at the second term of this summation, since any subsequent terms in the blue part of the summation above would only add to the total sum once you distribute the red minus sign.