The free groups can be totally (bi-)ordered. This paper shows how to do it (page 4). In short, you embed the group in multiplicative structure of the ring of power series in non-commuting variables, order those and transfer the ordering back to the free group via the inverse of the embedding.
Is it possible to explain this order intrinsically, without reference to power series in non-commuting variables? It would be enough for me to understand this order for the free group on two generators. In particular, what is the positive cone of this ordered group? In terms of the power series representation it's the series in which the first non-zero coefficient is negative. But how do you see this in a word on $\{x,y\}$?
There is a construction of a bi-ordering on $F_n$ due to Vinogradov:
A.A. Vinogradov. On the free product of ordered groups. Sbornik Mathematics 25 (1949), 163-168.
which is exactly along the lines you are asking. Moreover, he proves that any free product of bi-orderable groups $G_i$, $i\in I$, is bi-orderable. The construction is by induction on the word length and is a bit tedious, but explicit. It is explained for instance in section 2.1.2 of
"Groups, Orders, and Dynamics", by B. Deroin, A. Navas, C. Rivas.
The resulting order is different from the one defined by Magnus, so if you are interested in describing explicitly the order defined by Magnus it will not be useful.