The wall-crossing action on the Grothendieck group is multiplication by 1+s

The map $c$ is a well-defined homomorphism: by [F1] the classes of the standards are a $\mathbb{Z}$-basis indexed by the $\mathrm{\mid }W\mathrm{\mid }$ elements of $W$; sending the basis element $[\mathrm{\Delta }(v\cdot \lambda )]$ to $v$ therefore extends uniquely to an isomorphism of free abelian groups of rank $\mathrm{\mid }W\mathrm{\mid }$.

On every Verma class the formula holds: $c[{\theta }_s\mathrm{\Delta }(v\cdot \lambda )]=v+vs=v(1+s)=c[\mathrm{\Delta }(v\cdot \lambda )]\cdot (1+s)$ by [F2], using that right multiplication by $s$ in $\mathbb{Z}[W]$ sends the basis element $v$ to $vs$. Since ${\theta }_s$ is exact by [F2] it induces an additive endomorphism of $K_0$; agreement on the standard basis [F1] gives $c[{\theta }_{s}X]=c[X](1+s)$ for all $X\in {\mathcal{O}}_{\lambda }$, and both sides are additive in $X$.

Iterate step 2.1 along the word, applying the factors in the order in which they act: the composite ${\theta }_{s_k}\cdots {\theta }_{s_1}$ applies ${\theta }_{s_1}$ first, so $c[{\theta }_{s_k}\cdots {\theta }_{s_1}\mathrm{\Delta }(\lambda )]=(((1+s_1))(1+s_2))\cdots (1+s_k)=(1+s_1)(1+s_2)\cdots (1+s_k)$; expanding the product gives the sum over all subwords $s_{i_1}\cdots s_{i_r}$, $i_1<\cdots <i_r$, of their products, with the multiplicity with which each product arises. The full subword contributes $w$ once; every proper subword has strictly smaller length, and its product is $\le w$ with smaller length, hence strictly smaller than $w$ in the Bruhat order by [F3]. Reversing the order of composition multiplies the factors in the reverse order, as recorded.