Proof for Linearity of Expectation Question 
It is mentioned that the second line of the proof is from the definition of the Union of Probabilities: 

I do not understand how that happened. I understand that the first line is the definition of Expectation expanded, but, why are the two random variables being operated by an AND? 


I do not understand how that happened. I understand that the first line is the definition of Expectation expanded, but, why are the two random variables being operated by an AND? 

You have two variables, $X,Y$, and wish to perform a weighed sum over all outcomes in the samples space, $\Omega$.   We then partition the sample space by the joint supports of the random variables; thus we are weighting by the probabilities for the conjunction.
$$\begin{align}\text{In full:}\qquad\\\mathsf E(X+Y) ~&= \sum_{\omega\in\Omega}(X(\omega)+Y(\omega)) \,\mathsf P\{\omega\} \\ &= \sum_{i\in X(\Omega)}\sum_{j\in Y(\Omega)}\mathop{\sum\qquad\quad}_{\omega:(X(\omega)=i )\wedge(Y(\omega)=j)}(X(\omega)+Y(\omega))\, \mathsf P\{\omega\} \\ &= \sum_{i\in X(\Omega)}\sum_{j\in Y(\Omega)}(i+j)\mathop{\sum\qquad\quad}_{\omega:(X(\omega)=i )\wedge(Y(\omega)=j)} \mathsf P\{\omega\} \\ &= \sum_{i\in X(\Omega)}\sum_{j\in Y(\Omega)} (i+j) \,\mathsf P\{\omega:X(\omega)=i~\wedge~ Y(\omega)=j\} \\[3ex]\text{In short:}\quad\\\mathsf E(X+Y) ~ &=\sum_{i}\sum_{j} (i+j)\,\mathsf P(X=i\cap Y=j)\end{align}$$

Alternatively, you can view it as $\{X+Y=u\} = \bigcup_{i}\{X=i\}\cap\{Y=u-i\}$
So $\sum_{u}(u)\mathsf P\{X+Y=u\}~{=\sum_{u}(u)\sum_{i}\mathsf P\{X=i\}\cap\{ Y=u-i\}\\=\sum_i\sum_j (i+j)\mathsf P\{X=i\}\cap\{Y=j\}}$
