Let $p_0 = p_{2024} = 0$ and let $t$ be a positive integer such that $p_t = 2023$. The given condition gives
$$\begin{aligned} & 2 = 4048 - (2023 + 2023) \\ &= \left( p_1 + \sum_{i=1}^{2022} |p_{i+1} - p_i| + p_{2023} \right) - \left( \sum_{i=0}^{t-1} (p_{i+1} - p_i) + \sum_{i=t}^{2023} (p_i - p_{i+1}) \right) \\ &= \left( \sum_{i=0}^{t-1} |p_{i+1} - p_i| + \sum_{i=t}^{2023} |p_i - p_{i+1}| \right) - \left( \sum_{i=0}^{t-1} (p_{i+1} - p_i) + \sum_{i=t}^{2023} (p_i - p_{i+1}) \right) \\ &= \sum_{i=0}^{t-1} \left( |p_{i+1} - p_i| - (p_{i+1} - p_i) \right) + \sum_{i=t}^{2023} \left( |p_i - p_{i+1}| - (p_i - p_{i+1}) \right), \end{aligned}$$
thus
$$f(i) = \begin{cases} p_{i+1} - p_i & (0 \le i \le t-1), \\ p_i - p_{i+1} & (t \le i \le 2023) \end{cases}$$
satisfies $\sum_{i=0}^{2023} (|f(i)| - f(i)) = 2$. Since we have
$$|a| - a = \begin{cases} 0 & (a \ge 0), \\ 2|a| & (a < 0) \end{cases}$$
for any integer $a$, $\sum_{i=0}^{2023} (|f(i)| - f(i)) = 2$ if and only if there exists an integer $k$ with $0 \le k \le 2023$ such that $f(k) = -1$ and $f(i) > 0$ for every integer $i$ which is different from $k$ and $0 \le i \le 2023$. Since $f(0)$, $f(t-1)$, $f(t)$ and $f(2023)$ are all positive, $k$ is neither $0$, $t-1$, $t$ nor $2023$.

When $k \ge t+1$, we have $0 = p_0 < p_1 < \dots < p_t > p_{t+1} > \dots > p_k < p_{k+1} > p_{k+2} > \dots > p_{2024} = 0$ with $p_{k+1} = p_k + 1$. Since $p_{k+1} < 2023$, we have $p_k = p_{k+1} - 1 \le 2021$. Let $(A, B)$ be a pair of sets satisfying $A \cap B = \emptyset$, $A \cup B = \{1, 2, \dots, m-1, m+2, m+3, \dots, 2022\}$ for some integer $m$ with $1 \le m \le 2021$. We fix such a pair $(A, B)$, and consider all permutations $p_1, p_2, \dots, p_{2023}$ that satisfy the following conditions:
$$\bullet \{p_1, p_2, \dots, p_{t-1}\} = A,$$
$$\bullet p_t = 2023,\ p_k = m,\ p_{k+1} = m+1,$$
$$\bullet \{p_{t+1}, p_{t+2}, \dots, p_{k-1}, p_{k+2}, p_{k+3}, \dots, p_{2023}\} = B$$
The number of such permutations is to be determined. Note that $p_1, p_2, \dots, p_{t-1}$ are arranged in increasing order of elements in $A$. The numbers $p_{t+1}, p_{t+2}, \dots, p_{k-1}$ are arranged in decreasing order of the elements in $B$ that are greater than $m$, and the numbers $p_{k+2}, p_{k+3}, \dots, p_{2023}$ are arranged in decreasing order of the elements in $B$ that are less than $m+1$. Therefore, $p_{t+1}, p_{t+2}, \dots, p_{k-1}, p_{k+2}, p_{k+3}, \dots, p_{2023}$ are arranged in decreasing order of elements in $B$. Conversely, this permutation satisfies all the conditions, hence we conclude that there is exactly one such permutation. Since there are $2021$ choices for $m$, and $2^{2020}$ choices for $(A, B)$, there are $2021 \cdot 2^{2020}$ permutations satisfying the conditions when $k \ge t+1$.

By the same reasoning, the number of permutations satisfying the conditions when $k \le t-2$ is also $2021 \cdot 2^{2020}$. Thus, the answer is $2021 \cdot 2^{2020} \cdot 2 = 2021 \cdot 2^{2021}$.