Since I didn't want to think about permutations too much, here is an answer about relating Coxeter transformations of the form $C(A, \text{id})$ to Coxeter transformations of finite-dimensional path-algebras.

Let $A$ be a generalized Cartan matrix. Let $A_+$ and $A_-$ be defined by

$$\left( A_+\right)_{ij} =
\begin{cases}
A_{ji}, \ i > j \\
1, \ i = j \ \ \ \ \ \ \ \ ; \\
0, \text{else}
\end{cases}
\ \ \ (A_-)_{ij} =
\begin{cases}
A_{ij}, \ i > j \\
1, \ i = j \\
0, \text{else}
\end{cases}$$

**Theorem 1:** We have $C(A, \text{id}) = -A_{+}^{-1}A_{-}^t$.

*Proof:* This is Theorem $2.9$ in this paper by Sefi Ladkani. Note that I changed the definition of $A_+$ and $A_-$ to match the definition of a Coxeter transformation I gave in my question. $\square$

Now let $H = kQ$ be the path algebra of the finite quiver $Q$ without oriented cycles. Let $S(1), \dots, S(n)$ be the simple modules, ordered in such a way that there is never an arrow in increasing direction, i.e. whenever $1 \leq i \leq j \leq n$ then there is no arrow $i \to j$ in $Q$. This can be achieved by labeling a sink of $Q$ with number $1$, a sink of the remaining quiver after killing $1$ with number $2$ and so on. The homological bilinear form $\left\langle - , -\right\rangle : \mathbb{Z}^n \times \mathbb{Z}^n \to \mathbb{Z}$ is given (since $H$ is hereditary) on dimension vectors by
\begin{align*}
\left\langle \underline{\dim} S(i), \underline{\dim} S(j)\right\rangle
& = \dim_k \text{Hom}_H(S(i), S(j)) - \dim_k \text{Ext}_H^1(S(i), S(j)) \\
& = \delta_{ij} - \#\{\alpha: i \to j\}.
\end{align*}
It is well known that the Coxeter transformation $\Phi_H$ of $H$ is the unique Coxeter transformation of the homological bilinear form, in the sence of Ladkanis paper (i.e. $\left\langle x,y \right\rangle = - \left\langle y, \Phi_{H}x \right\rangle$). Therefore, if we set $D$ the matrix with entries $(D)_{ij} = \left\langle \underline{\dim}S(i), \underline{\dim}S(j)\right\rangle$ then we get $\Phi_H = -D^{-1}D^t$. Furthermore, the matrix $A = D + D^t$ is a symmetric generalized Cartan matrix, and since $D$ has no nonzero entries at the upper triangle (remember the order of the simple modules), we get $A_+ = D = A_-$. We get the following:

**Theorem 2:** $\Phi_H = C(A, \text{id})$.

*Proof:* By theorem $1$ we have $C(A, \text{id}) = -A_+^{-1}A_{-}^t = -D^{-1}D^t = \Phi_H$. Compare also with Corollary $2.11$ in Ladkanis paper. $\square$

*Remark:* I'm pretty sure we can generalize this to every permutation. Also note that we can in this way represent every Coxeter transformation of a *symmetric* generalized Cartan matrix as the Coxeter transformation of a path algebra (since the numbers in the lower triangle show us exactly how many arrows we have to put between the indices).

**Added later**: There is still the question how properties of the algebra $H = kQ$ and properties of the associated Cartan matrix $A$ correspond. Remember that we have $A_{ii} = 2$, $A_{ij} = - \#\{\alpha: i \to j\}$ for $i > j$ and $A_{ij} = A_{ji}$ and that al arrows in $Q$ go in increasing direction. Let $q_A$ be the quadratic form associated to $A$, i.e. $q_A(x) = x^{t}Ax$ for $x \in \mathbb{Z}^n$. Let $q_Q$ be the quadratic form of the quiver $Q$, i.e.
$$q_Q(x) = \left\langle x, x \right\rangle = \sum_{i = 1}^{n}x_i^2 - \sum_{\alpha \in Q_1}x_{s(\alpha)}x_{t(\alpha)}.$$

**Theorem 3:** $q_A = 2q_Q$.

*Proof:* We have
\begin{align*}
q_A(x) & = x^tAx = \sum_{i,j} x_iA_{ij}x_j \\
& = \sum_{i = 1}^{n}A_{ii}x_{i}^2 + \sum_{i \neq j} A_{ij}x_ix_j \\
& = \sum_{i = 1}^{n}2x_i^2 + \sum_{i > j}2A_{ij}x_ix_j \\
& = 2 \left( \sum_{i = 1}^{n}x_i^2 - \sum_{\alpha \in Q_1} x_{s(\alpha)}x_{t(\alpha)}\right) \\
& = 2q_Q(x),
\end{align*}
proving the claim. $\square$

Therefore, the quiver $Q$ is Dynkin (i.e. $H$ is representation finite), Euclidean (i.e. $H$ is tame) or wild (i.e. $H$ is wild) iff $A$ is positive definite, positive semidefinite (but not positive definite) or indefinite as a matrix.