My understanding is the following. They want to determine the next state of the order book. So, they need to know which order comes first.

One way to find which order arrives first could be to generate a new $X_i$ for all $i \in \mathcal{I}$, and to identify the smallest. This would not be efficient. The alternative that they mention relies on $$ p_k := {\rm Pr}[\min (X_i : i \in \mathcal{I}) = X_k] = \frac{\lambda_k}{\sum_{i \in \mathcal{I}} \lambda_i}, $$ which provides the probability that the $k$th is the smallest. Obviously, these probabilities sum to one. The idea is to partition the interval $[0, 1]$ into disjoint segments of size $p_k$, one for each order $k$. To determine which order comes first, one just needs to sample a uniform random variable on $[0,1]$, and determine in which segment it belongs.

Edit: example of determination of the first order
Suppose that $\mathcal{I} = \{1, 2, 3\}$, and that $p_1 = 0.2, p_2 = 0.5, p_3 = 0.3$. Then the segments are [0, 0.2], (0.2, 0.7] and (0.7, 1]. Let $u$ denote the value sampled from the uniform distribution. If $u \in [0, 0.2]$ then $k = 1$, if $u \in (0.2, 0.7]$ then $k = 2$, and if $u \in (0.7, 1]$ then $k = 3$.

My understanding is the following. They want to determine the next state of the order book. So, they need to know which order comes first.

One way to find which order arrives first could be to generate a new $X_i$ for all $i \in \mathcal{I}$, and to identify the smallest. This would not be efficient. The alternative that they mention relies on $$ p_k := {\rm Pr}[\min (X_i : i \in \mathcal{I}) = X_k] = \frac{\lambda_k}{\sum_{i \in \mathcal{I}} \lambda_i}, $$ which provides the probability that the $k$th is the smallest. Obviously, these probabilities sum to one. The idea is to partition the interval $[0, 1]$ into disjoint segments of size $p_k$, one for each order $k$. To determine which order comes first, one just needs to sample a uniform random variable on $[0,1]$, and determine in which segment it belongs.

Edit: example of determination of the first order
Suppose that $\mathcal{I} = \{1, 2, 3\}$, and $\lambda_1 = 2, \lambda_2 = 5, \lambda_3 = 3$. So, $p_1 = 0.2, p_2 = 0.5, p_3 = 0.3$, and the segments are [0, 0.2], (0.2, 0.7] and (0.7, 1]. 
Let $u$ denote the value sampled from the uniform distribution. If $u \in [0, 0.2]$ then $k = 1$, if $u \in (0.2, 0.7]$ then $k = 2$, and if $u \in (0.7, 1]$ then $k = 3$.

My understanding is the following. They want to determine the next state of the order book. So, they need to know which order comes first.

One way to find which order arrives first could be to generate a new $X_i$ for all $i \in \mathcal{I}$, and to identify the smallest. This would not be efficient. The alternative that they mention relies on $$ p_k := {\rm Pr}[\min (X_i : i \in \mathcal{I}) = X_k] = \frac{\lambda_k}{\sum_{i \in \mathcal{I}} \lambda_i}, $$ which provides the probability that the $k$th is the smallest. Obviously, these probabilities sum to one. The idea is to partition the interval $[0, 1]$ into disjoint segments of size $p_k$, one for each order $k$. To determine which order comes first, one just needs to sample a uniform random variable on $[0,1]$, and determine in which segment it belongs.

My understanding is the following. They want to determine the next state of the order book. So, they need to know which order comes first.

One way to find which order arrives first could be to generate a new $X_i$ for all $i \in \mathcal{I}$, and to identify the smallest. This would not be efficient. The alternative that they mention relies on $$ p_k := {\rm Pr}[\min (X_i : i \in \mathcal{I}) = X_k] = \frac{\lambda_k}{\sum_{i \in \mathcal{I}} \lambda_i}, $$ which provides the probability that the $k$th is the smallest. Obviously, these probabilities sum to one. The idea is to partition the interval $[0, 1]$ into disjoint segments of size $p_k$, one for each order $k$. To determine which order comes first, one just needs to sample a uniform random variable on $[0,1]$, and determine in which segment it belongs.

Edit: example of determination of the first order
Suppose that $\mathcal{I} = \{1, 2, 3\}$, and that $p_1 = 0.2, p_2 = 0.5, p_3 = 0.3$. Then the segments are [0, 0.2], (0.2, 0.7] and (0.7, 1]. Let $u$ denote the value sampled from the uniform distribution. If $u \in [0, 0.2]$ then $k = 1$, if $u \in (0.2, 0.7]$ then $k = 2$, and if $u \in (0.7, 1]$ then $k = 3$.