Well, it seems that the question is still unanswered. In general, there are different ways of encoding, but -- as I understand it -- you are asking for a unitary way of doing it.
Suppose we have a $[[n,k]]$ stabiliser code $C$ with generators $g_1,\dots,g_{n-k}$.
This code is Clifford-equivalent to the code $\mathsf{Z}_k:=|0^{n-k}\rangle\otimes (\mathbb{C}^2)^{\otimes k}$ stabilised by $Z_{1},\dots, Z_{n-k}$.
The encoding construction is given by a choice of Clifford unitary mapping $\mathsf{Z}_k$ to $C$.
In general, this Clifford unitary $U$ is not unique since we can redefine it by any Clifford $V$ which acts trivially on $\mathsf{Z}_k$ (or $C$ respectively).
However, any choice of stabiliser basis of $C$ induces a $U$. This is simply because a choice of logical basis $|\bar x\rangle$ corresponds to a maximal completion $g_1,\dots,g_{n-k},g_{n-k+1},\dots,g_n$ of the generators and the logical basis is determined by the eigenvalues of $g_{n-k+1},\dots,g_n$. The Clifford unitary $U$ is defined by the following equations:
$$
U Z_i U^\dagger = g_i, \qquad \forall i = 1,\dots,n.
$$
Since the Paulis on the left and right hand side are maximally commuting sets, this indeed defines a Clifford unitary $U$. It is straightforward to check that any input state is transformed as
$$
|0^{n-k}\rangle\otimes |\psi\rangle = \sum_{x\in\mathbb F_2^{k}} \psi_x |0^{n-k}\rangle\otimes |x\rangle \stackrel{U}{\longmapsto} \sum_{x\in\mathbb F_2^{k}} \psi_x |0^{n-k}\rangle\otimes |\bar{x}\rangle.
$$
Final remarks:
- The binary/symplectic description of the global Clifford unitary $U$ is evident from the definition.
- This can be used to compile $U$ into generators via standard methods.