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It does not matter. The start time and the end time of intervals are in equal position for this problem.
e.g.
[1, 5], [2, 10], [6, 9] is currently sorted by start. You may merge them from left to right and get the result [1,10].
If they were sorted by end time instead - [1, 5], [6, 9],[2, 10]. You could merge them from right to left and get the same result [1, 10].

It just depends on if you wanna start the merge from the left or right side of the given inteval list.
A sample code for when intervals sorted by start time

public List<Interval> merge(List<Interval> intervals) {
    if (intervals.size() <= 1)
        return intervals;
    
    // Sort by ascending starting point 
    intervals.sort((i1, i2) -> Integer.compare(i1.start, i2.start));
    
    List<Interval> result = new LinkedList<Interval>();
    int start = intervals.get(0).start;
    int end = intervals.get(0).end;
    
    for (Interval interval : intervals) {
        if (interval.start <= end) // Overlapping intervals, move the end if needed
            end = Math.max(end, interval.end);
        else {                     // Disjoint intervals, add the previous one and reset bounds
            result.add(new Interval(start, end));
            start = interval.start;
            end = interval.end;
        }
    }
    
    // Add the last interval
    result.add(new Interval(start, end));
    return result;
}

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depends what you mean exactly by "merge", e.g.
a: [1..7]
b: [5..10]
c: [9..13]
and you want [1..13] as a merged interval of a,b,c
you might want to sort by start times if you want to merge starting by start. You walk through the sorted intervals, pick the first and if the next overlaps you max on the end until no further overlap with maxed end occurs...

If you like thinking backwards, just do the same from the end with minimized start. Its symetric... so it seems to not matter on the result and run time.

If you want to work a bit more with your intervals, it may make sense to put them into an interval tree. you may google it.