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 $F_2$. 

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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. 

<cite authors="Linnell, Peter A.; Rhemtulla, Akbar; Rolfsen, Dale P. O.">_Linnell, Peter A.; Rhemtulla, Akbar; Rolfsen, Dale P. O._, [**Discretely ordered groups**](http://dx.doi.org/10.2140/ant.2009.3.797), Algebra Number Theory 3, No. 7, 797-807 (2009). [ZBL1229.06008](https://zbmath.org/?q=an:1229.06008).</cite>