I know that for Steane code, we can implement transversally some gates like cNOT, Hadamard and Pauli.
What I am looking for is a resource in which it is explained why implementing those gate give rise to the good logical operation.
If this is a result more general than Steane I would be interested in a "general enough" approach showing it.
Let $\mathcal{H}$ be the Hilbert space of a set of physical qubits and let $S$ be the stabilizer group of a stabilizer code $\mathcal{G} \subset \mathcal{H}$.
A transversal operator $U$ on $\mathcal{H}$ implements a logical operator on $\mathcal{G}$ if it maps $\mathcal{G}$ back to itself. This can be established by showing that $U$ does not change the stabilizer group $S$ of $\mathcal{G}$. In order to do this, we need to know how stabilizers transform under unitary gates. Suppose that $g$ is an operator that stabilizes $|\psi\rangle$, i.e. $g|\psi\rangle = |\psi\rangle$. Then
$$ UgU^\dagger U|\psi\rangle = Ug|\psi\rangle = U|\psi\rangle\tag1 $$
and we see that $UgU^\dagger$ stabilizes $U|\psi\rangle$. We will use this fact to demonstrate that certain transversal operators do not change the stabilizer group of stabilizer codes that meet certain criteria. We will also use $(1)$ to determine the action of those transversal operators within the code subspace by analyzing their effect on the logical Pauli operators.
Claim: If $\mathcal{G}=CSS(C_1, C_2)$ is a Calderbank-Shor-Steane code for classical linear codes $C_1$ and $C_2$, $C_2^\perp \subset C_1$ then transversal CNOT is a logical CNOT on $\mathcal{G}$.
Remark 1: There is ambiguity in the use of the $CSS(C_1, C_2)$ notation in the literature. For example, Wikipedia and Nielsen & Chuang put $C_2^\perp$ in the second position rather than $C_2$.
Remark 2: The condition $C_2^\perp \subset C_1$ is not an additional restriction on $\mathcal{G}$. It is part of the definition of a CSS code necessary to make sure that the $X$ type and $Z$ type stabilizer generators commute. Consequently, the claim says that CNOT admits transversal implementation for any CSS code.
Proof sketch: Let $g_x$ be a tensor product of identity $I$ and Pauli $X$ operators and similarly for $g_z$. Since $\mathcal{G}$ is a CSS code, we can choose stabilizer generators that are either of the form $g_x$ or of the form $g_z$. Calculate that
$$ \mathrm{CNOT} \circ (g_x \otimes I) \circ \mathrm{CNOT} = g_x \otimes g_x \\ \mathrm{CNOT} \circ (I \otimes g_x) \circ \mathrm{CNOT} = I \otimes g_x \\ \mathrm{CNOT} \circ (g_z \otimes I) \circ \mathrm{CNOT} = g_z \otimes I \\ \mathrm{CNOT} \circ (I \otimes g_z) \circ \mathrm{CNOT} = g_z \otimes g_z. $$
Note that if $g_x, g_z \in S$ then all operators on the right hand sides of the four equations above are in $S \times S$. Moreover, every operator in $S \times S$ can be obtained as a composition of operators of the form $g_x \otimes I$, $g_z \otimes I$, $I \otimes g_x$ and $I \otimes g_z$. We conclude that transversal CNOT preserves $S \times S$.
By analyzing the action of transversal CNOT on the logical $X$ and $Z$ operators along the same lines as above, we see that the logical operator effected by performing transversal CNOT on the physical qubits is in fact the logical CNOT. $\square$
Claim: If $\mathcal{G}=CSS(C_1, C_2)$ is a Calderbank-Shor-Steane code where the two classical linear codes are the same $C_1 = C_2$ then transversal Hadamard is a logical Hadamard on $\mathcal{G}$.
Remark: The definition of the CSS code requires that $C_2^\perp \subset C_1$, so $C_1$ cannot be arbitrary. It necessarily contains its own dual.
Proof sketch: As before, let $g_x$ be a tensor product of identity $I$ and Pauli $X$ operators. We see that
$$ H g_x H = g_z \\ H g_z H = g_x. $$
where $g_z$ is obtained from $g_x$ by replacing $X$ operators in the tensor product with $Z$ operators. Note that since $\mathcal{G}$ is a CSS code, we can choose stabilizer generators that are either of the form $g_x$ or of the form $g_z$. Moreover, since $C_1 = C_2$, $g_x \in S$ if and only if $g_z \in S$. Consequently, transversal Hadamard preserves the stabilizer.
As before, the analysis extends to logical $X$ and $Z$ operators and therefore transversal $H$ on physical qubits implements the logical Hadamard gate. $\square$
(For completeness we include the phase gate even though it is not mentioned in the question.)
Claim: If $\mathcal{G}=CSS(C_1, C_2)$ is a Calderbank-Shor-Steane code where the two classical linear codes are the same $C_1 = C_2$ and $C_2^\perp\subset C_1$ is doubly-even (i.e. all codewords in $C_2^\perp$ have Hamming weight divisible by four) then transversal phase gate $P$ preserves the stabilizer $S$.
Proof sketch: Since the phase gate commutes with the Pauli $Z$ operator it is clear that all $g_z$ stabilizers are preserved. For any $g_x$, we have
$$ P g_x P^\dagger = i^{w(g_x)} g_x g_z $$
where $w(g_x)$ is the weight of stabilizer $g_x$, i.e. the number of non-identity factors in it. Since $C_2^\perp$ is doubly-even, we see that
$$ P g_x P^\dagger = g_x g_z. $$
Finally, since $C_1=C_2$ we have $g_z\in S$ and thus $P g_x P^\dagger\in S$. $\square$
Note that the transversal phase operator is not necessarily the logical phase gate. However, $C_2^\perp$ is an even code so the transversal $Z$ belongs to the normalizer $N(S)$. Thus, as long as transversal $Z$ is not in the stabilizer we can choose it to play the role of the logical $Z$. Under this choice transversal phase operator commutes with the logical $Z$ and therefore its action on $\mathcal{G}$ is a diagonal gate. Moreover, applying transversal phase operator twice yields the logical $Z$. We conclude that in this case the transversal phase operator is either a logical phase gate or its inverse.
Claim: If $\mathcal{G}$ is a stabilizer code, then logical Pauli operators are transversal.
Proof sketch: This follows immediately from the fact that logical Pauli operators are chosen from the normalizer $N(S)$ of the stabilizer group $S$ in the $n$-qubit Pauli group $\mathcal{P}_n$ since all operators in $\mathcal{P}_n$ are transversal by definition. $\square$
Steane code is $CSS(C_1, C_2)$ where $C_1 = C_2$ is the Hamming $[7, 4, 3]$ code.
Since it is a stabilizer code, Pauli operators are transversal. Since it is a CSS code, CNOT is transversal. Since $C_1 = C_2$, Hadamard is transversal. Since $C_1$ is also doubly-even, the phase gate is transversal.
Thus, the entire Clifford group admits a transversal implementation in the Steane code. By Eastin-Knill theorem, we cannot extend the set of transversal gates to a universal gateset. In particular, the $T$ gate does not have a transversal implementation in the Steane code.
