Take n elements. If at end start from begining

You don't need to track the overflow because we can use the % modulus operator (which returns the remainder from an integer division) to continually loop through a range of indexes, and it will always return a valid index in the collection, wrapping back to 0 when it gets to the end (and this will work for multiple wraps around the end of the list):

List<int> list = new List<int> {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
List<int> newList = new List<int>();

for (int skip = 9, take = 2; take > 0; skip++, take--)
{
    newList.Add(list[skip % list.Count]);
}

Result:

// newList == { 10, 1 }

---

This could be extracted into an extension method:

public static List<T> SkipTakeWrap<T>(this List<T> source, int skip, int take)
{
    var newList = new List<T>();

    while (take > 0)
    {
        newList.Add(source[skip % source.Count]);
        skip++;
        take--;
    }

    return newList;
}

And then it could be called like:

List<int> list = new List<int> { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
List<int> newList = list.SkipTakeWrap(9, 2);

you may need to do something like this

        var start = 9;
        var amount = 2;
        List<int> list = new List<int>() { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
        List<int> listOverflow = list.ToList();

        var overflow = (start + amount) - list.Count;
        if (overflow > 0)
            for (var i = 0; i < overflow; i++)
                listOverflow.AddRange(list.ToList());

        var newList = listOverflow.Skip(start).Take(amount).ToList();

use an extension method to circular repeat the enumerable

public static IEnumerable<T> Circular<T>( this IEnumerable<T> source )
{
    while (true)
        foreach (var item in source) 
            yield return item;
}

and you can use your code

List<int> list = new List<int>() {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
List<int> newList = list.Circular().Skip(9).Take(2).ToList();

.net fiddle example