Total ordering on the free group 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 $\{x,y\}$? 
 A: 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.  
A: It is possible to give a formal definition of the total order in terms of combinatorics on words (although it is merely a direct translation of the definition using the Magnus automorphism). It relies on the extension to words of the binomial coefficients (see below).
Let $A$ be a finite, totally ordered alphabet, for instance $A = \{a, b\}$ with $a < b$. Let $(x_n)_{n \geqslant 0}$ be the sequence of words of the free monoid $A^*$ (totally) ordered by the shortlex order: $$1 < a < b < aa < ab < ba < bb < aaa < \dotsm$$

Short answer. The free group $F(A)$ is totally (bi-)ordered by setting, for $u, v \in F(A)$,
$$   u < v \text{ if and only if, there exists $n \geqslant 0$ such
 that $\binom{u}{x_k} = \binom{v}{x_k}$ for $0 \leqslant k < n$ and $\binom{u}{x_n}
 < \binom{v}{x_n}$}. $$

Let me recall some useful definitions. 
Shortlex order.  The lexicographic order $\leqslant_{lex}$ on the free monoid $A^*$ is the order used in a dictionary. The shortlex order on  $A^*$ is defined as follows. If $u, v \in A^*$, then $u \leqslant v$ if either $|u| < |v|$ or $|u| = |v|$ and $u \leqslant_{lex} v$.
Subword. A word $u$ is a subword of $v$ if $v$ can be written as 
$v = v_0u_1v_1u_2v_2\dotsm u_kv_k$
where $u_i$ and $v_i$ are (possibly empty) words such that $u_1u_2\dotsm u_k = u$. For instance, the words $\color{red}{baba}$ and $\color{red}{acab}$ are subwords of $abcacbab$ since
$abcacbab = a\color{red}{b}c\color{red}{a}c\color{red}{ba}b = \color{red}{a}b\color{red}{ca}c\color{red}{b}ab$.
Binomial coefficients of words in $A^*$. Given two words $u$ and $v$, let $\binom{v}{u}$ denote the number of distinct ways to write $u$ as a subword of $v$.
More formally, if $u = a_1a_2 \cdots a_n$, then
$$
 \binom{v}{u} = \left| \left\{ (v_0, \ldots,v_n) \mid v
 = v_0a_1v_1 \ldots a_nv_n \right\} \right|.
$$
Observe that if $u$ is a letter $a$, then $\binom{v}{a}$ is simply the number of
occurrences of the letter $a$ in $v$. Also note
that if $A = \{a\}$, $u = a^n$ and $v = a^m$, then
$$
 \binom{v}{u} = \binom{m}{n}
$$ 
and hence these numbers constitute a generalization of the classical
binomial coefficients. The generalized binomial coefficients can be iteratively computed using the following generalisation of Pascal's triangle. Let $1$ be the empty word. Let $u,v \in A^*$ and $a,b \in A$. Then


*

*$\binom{u}{1} = 1$,

*$\binom{u}{v} = 0$ if $|u| \leqslant |v|$ and $u \neq v$,

*$\binom{ua}{vb} = \begin{cases}
  \binom{u}{vb}                &\text{if $a\not= b$}\\
  \binom{u}{vb} + \binom{u}{v} &\text{if $a= b$}
 \end{cases}$


Binomial coefficients $\binom{v}{x}$ where $v \in F(A)$ and $x \in A^*$.
One can extend the "Pascal's triangle" by setting, for $v \in F(A)$, $x \in A^*$ and $a \in A$
$$
 \binom{va^{-1}}{xa^{r}} = \sum_{0 \leq i < r} (-1)^i \binom{v}{xa^{r-i}} \text{
 where $x$ does not end with an $a$.}
$$
For instance, 
\begin{align*}
 \binom{aba^{-1}b^{-1}}{aba} &= \binom{aba^{-1}}{aba} = \binom{ab}{aba} -
 \binom{ab}{ab} = 0 - 1 = -1 \\
 \binom{aba^{-1}b^{-1}}{baa} &= \binom{aba^{-1}}{baa}  
 = \binom{ab}{baa}  - \binom{ab}{ba} + \binom{ab}{b} = 0 - 0 + 1 = 1 
\end{align*}
Connection with the Magnus automorphism. Let ${\mathbb Z}\langle\!\langle A\rangle\!\rangle$ be the algebra of noncommutative power series with coefficients in $\mathbb Z$
and variables in $A$. The Magnus transformation is the monoid morphism $\mu$
from the free monoid $A^*$ to ${\mathbb Z}\langle\!\langle A\rangle\!\rangle$ defined, for each letter $a \in A$, by $\mu(a)
= 1 + a$. Then for each $u \in A^*$,
\begin{equation}
 \mu(u) = \sum_{x \in A^*} \binom{u}{x} x 
\end{equation}
For instance 
\begin{align*}
  \mu(abaa) &= (1+a)(1+b)(1+a)(1+a) \\
  &= 1 + 3a + b  + 2aa + ab + 2ba + aaa + 2aba + baa + abaa
\end{align*}
Since the series $1 + a$ are invertible in ${\mathbb Z}\langle\!\langle A\rangle\!\rangle$, the
Magnus transformation can be extended to a group homomorphism from $F(A)$ to ${\mathbb Z}\langle\!\langle A\rangle\!\rangle$.
For instance, 
\begin{align*}
  \mu(a^{-1}) &= (1+a)^{-1} = \sum_{n \geq 0} (-1)^n a^n \\
 \mu(aba^{-1}b^{-1}) &= (1+a)(1+b)(1+a)^{-1}(1+b)^{-1} \\
 &= (1+a+b+ab)\Bigl(\sum_{n \geqslant 0} (-1)^n a^n\Bigr) \Bigl(\sum_{n \geq 0} (-1)^n 
 b^n \Bigr) \\
 &= 1 + ab - ba - aba - abb + baa + bab + \dotsm
\end{align*}
One can now define the binomial coefficient
$\binom{u}{x}$ for $u \in F(A)$ and $x \in A^*$ by setting
for each $u \in F(A)$,
\begin{equation}
 \mu(u) = \sum_{x \in A^*} \binom{u}{x} x 
\end{equation} 
For instance, one recovers the values
$$
\binom{aba^{-1}b^{-1}}{aba} = -1 \qquad \binom{aba^{-1}b^{-1}}{baa} = 1.
$$
