The Schreier coset graph is connected and deterministic

Let $K\le F(X)$ be the subgroup generated by the image of $X$. The inclusion $X\hookrightarrow K$ extends, by [L1], to a homomorphism $\varphi :F(X)\to K$, and the inclusion $K\hookrightarrow F(X)$ composed with $\varphi$ agrees with the identity of $F(X)$ on $X$. Uniqueness in [L1] therefore forces $(K\hookrightarrow F(X))\circ \varphi ={\mathrm{id}}_{F(X)}$, so $K=F(X)$. Thus every element of $F(X)$ is represented by a word on $X\bigsqcup X^{-1}$.

Let $Hg$ be any vertex. Choose a word $a_1\cdots a_n$ on $X\bigsqcup X^{-1}$ that represents $g$, and delete adjacent inverse pairs until the word is reduced. Reading the remaining letters from the base vertex $H$ follows the edges of [L3] forward for letters in $X$ and backward for letters in $X^{-1}$, and after the first $j$ letters one is at the coset $H{a}_1\cdots a_j$. The final vertex is therefore $Hg$, so the graph is connected.

For fixed $Hg$ and $x\in X$, [L3] gives exactly one outgoing $x$-edge, namely $Hg\to Hgx$. The same edge starts at $Hg{x}^{-1}$ and ends at $Hg$, so it is also the unique incoming $x$-edge into $Hg$. This is the required determinism.