It is not true that the crossing is necessarily in degree $1$. More precisely, if $\delta_{k,\mathbf i}$ is the crossing between strands $k$ and $k+1$ for the sequence $\mathbf i=(i_1,\ldots, i_m)$, then $$\deg(\delta_{k,\mathbf i})=-i_k\cdot i_{k+1}=\begin{cases}-2&\text{ if }i_k=i_{k+1},\\ 1 &\text{ if } i_k\text{ is connected to }i_{k+1}\text{ in the graph }\Gamma,\\ 0&\text{ otherwise.}\end{cases}.$$
In particular, in the case $i_k=i_{k+1}$ (such as in $\S$ 2.2 (3), which given your question title may be the case you are principally interested in), $\delta_{k,\mathbf i}$ indeed acts as a divided difference operator, hence lowers the degree of the polynomial by $2$, hence indeed has degree $-2$.
If $i_k\neq i_{k+1}$, then it is more complicated: now $\delta_{k,\mathbf i}$ does not act as a divided difference, but instead acts as a (possibly scaled) permutation of variables (see the description before Proposition 2.3). In particular, as noted shortly before Corollary 2.6, the grading on $\mathcal{Pol}_{\nu}=\bigoplus_{|\mathbf i|=\nu} \mathcal{Pol}_{\mathbf i}$ is determined by fixing a particular $\mathbf i\in{\rm Seq}(\nu)$ and fixing
$1_{\mathbf i}=1\in \mathcal{Pol}_{\mathbf i}$ to be in degree 0. The grading on $\mathcal{Pol}_{\nu}$ is the induced from this choice; in particular, the other units $1_{\mathbf j}$ may have non-zero degrees as elements of $\mathcal{Pol}_{\nu}$.
For example, consider $\S$ 2.2 (7): $\nu=i+j$ with $i\cdot j=-1$. Assume without loss of generality that the edge in $\Gamma$ is oriented $i\leftarrow j$. Then $\mathcal{Pol}_\nu=\mathcal{Pol}_{i,j}\oplus \mathcal{Pol}_{j,i}$. Let us set $\deg(1_{i,j})=0$. Then the unique crossing for $(i,j)$, $\delta_{1,(i,j)}=\delta_{i,j}$ acts by mapping e.g. $1_{i,j}\mapsto 1_{j,i}$, so in particular $\deg(1_{j,i})=1$ by definition. This is consistent because the reverse crossing $\delta_{j,i}$ maps $1_{j,i}\mapsto (x_1+x_2)1_{i,j}$ which has degree $2$ as expected.
