Let us first clarify the relationship between scattered, discrete, and dense orders. The last two notions are standard in the theory of ordered groups.
An order (left or two-sided) is discrete if there exists a smallest (necessarily unique) positive element. If such an element does not exist, the order is called dense.
We claim that no dense order (left or two-sided) is scattered. Moreover, we claim that, if the order is dense, then the entire group is a densely ordered set. Indeed, if $a < b$ then $e < a^{-1}b$ and, since the order is dense, there exists $c$ such that $e < c < a^{-1}b$ and we obtain $a < ac < b$.
On the other hand, discrete orders are not necessarily scattered. For instance the lexicographic bi-order on $\mathbb{Z} \times F_2$, with any bi-order on $F_2$, is discrete with smallest element $(1,e)$, but there are densely ordered subsets, namely the copy of $F_2$.
Going back to the posed question, since free groups of rank > 1 do not admit discrete bi-orders (in fact, no centerless group admits a discrete bi-order) they do not admit scattered bi-orders.
On the other hand, there are discrete left-orders on free groups of rank > 1, and, therefore, the question is more interesting there.
The claim that centerless groups do not admit discrete bi-orders follows from Theorem 2.1 in the paper cited below, while the claim that free groups admit discrete orders (that are, moreover Conradian) is Corollary 3.6 in the same paper.
Linnell, Peter A.; Rhemtulla, Akbar; Rolfsen, Dale P. O., Discretely ordered groups, Algebra Number Theory 3, No. 7, 797-807 (2009). ZBL1229.06008.