First note that the the comma in the proability is the AND operator; $$ \Pr[X = x , Y = y] = \Pr[X = x \wedge Y = y]$$ This is common notation to simplify the writing.
Now, explicitly write as
$$p_i = \sum_{j=1}^{n} r_{ij} = \Pr[X = x_i \wedge Y = y_0] + \Pr[X = x_i \wedge Y = y_1] + \cdots + \Pr[X = x_i \wedge Y = y_m]$$
Since the random variables $X$ and $Y$ are independent then this is just a partition of the event $x_i$ by the random variable $Y$.
As a solid case, consider two dices; one has the $X$ and the other is the $Y$ as their random variable representing the upper value of the dice. In total there are 36 possible equal values of the two dice roll. Fix the first one, let say $3$ then
\begin{align}\Pr(X=3) = & \Pr(X=3,Y=1)+\\
& \Pr(X=3,Y=2)+\\
& \Pr(X=3,Y=3)+\\
& \Pr(X=3,Y=4)+\\
& \Pr(X=3,Y=5)+\\
& \Pr(X=3,Y=6)\\
= &\frac{1}{36}+ \frac{1}{36}+ \frac{1}{36}+ \frac{1}{36}+ \frac{1}{36} +\frac{1}{36} = \frac{1}{6}
\end{align}
$H(X,Y)$ is actually the Joint Entropy and the formula is given by (again the AND);
$$H(X,Y) = -\sum_{x\in\mathcal X} \sum_{y\in\mathcal Y} P(x,y) \log_2[P(x,y)]$$
In our context this is
$$H(X,Y) = -\sum_{x\in X} \sum_{y\in Y} P(X=x,Y=y) \log_2[P(X=x,Y=y)]$$
$H(X,Y)$ is simultaneous evaluation of $X$ and $Y$ and that is equal to first evaluating $X$ then given value of $X$ evaluate the $Y$
$$H(X,Y)= H(X|Y)+H(Y)=H(Y|X)+H(X) $$
Proving this bit long;
\begin{align}
H(X,Y) & = − \sum_{i=1}^n \sum_{j=1}^m \Pr(X=x_i,Y =y_j) \log \big( \Pr(X=x_i,Y =y_j) \big)\\
& = − \sum_{i=1}^n \sum_{j=1}^m \Pr(X=x_i,Y =y_j) \log \big( \Pr(X=x_i) \Pr(Y|X = y_j|x_i) \big)\\
& = − \sum_{i=1}^n \sum_{j=1}^m \Pr(X=x_i,Y =y_j) \big[ \log \big( \Pr(X=x_i) \big) + \log \big( \Pr(Y|X = y_j|x_i) \big) \big] \\
& = − \sum_{i=1}^n \left( \sum_{j=1}^m \Pr(X=x_i,Y =y_j) \right) \log \big( \Pr(X=x_i) \big) \\
& - \sum_{i=1}^n \sum_{j=1}^m \Pr(X=x_i,Y =y_j) \log \left( \Pr(Y|X = y_j|x_i) \right)\\
& = H(X) + H(Y|X)
\end{align}