I was able to reduce this time complexity from O(n) = 2n to O(n) = n by using a list for the elements to be reversed, and then only modifying those rather than stepping through all nodes while reversing.

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Solution

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseBetween(self, head: Optional[ListNode], left: int, right: int) -> Optional[ListNode]:
        first = head
        start = None
        end = None
        last = None
        toReverse = []
        i = 0
        while i < right:
            if i == max(left - 2, 0):
                start = head
            if i == right - 1:
                end = head
            if i >= left - 1 and i < right - 1:
                toReverse.append(head)
            head = head.next
            i += 1
        
        last = end.next
        if left > 1:
            start.next = end
            start = start.next
        else:
            start = end
            first = start
        head = start
        while toReverse:
            head.next = toReverse.pop()
            head = head.next
        if head.next:
            head.next = last

        return first