Curve velocity depends only on the contact class

Choose a chart $(U,x)$ witnessing contact equivalence, write $a:=x(p)$ and ${\alpha }_r:=x\circ {\gamma }_r$, and represent the germ $[f]$ by the smooth function $\tilde{f}:=f\circ x^{-1}$ near $a$. By [L1], after shrinking if needed there are smooth functions $g_1,\dots ,g_n$ near $a$ such that $\tilde{f}(u)-\tilde{f}(a)={\sum }_{i=1}^n(u^i-a^i)g_i(u)$ and $g_i(a)={\mathrm{\partial }}_i\tilde{f}(a)$.

Substituting $u={\alpha }_r(t)$ into the identity from step 1.1, dividing by $t$, and letting $t\to 0$ gives $(f\circ {\gamma }_r{)}^{\mathrm{\prime }}(0)=(\tilde{f}\circ {\alpha }_r{)}^{\mathrm{\prime }}(0)={\sum }_{i=1}^n{g}_i(a)({\alpha }_r^i{)}^{\mathrm{\prime }}(0)\mathrm{.}$ Because contact equivalence means ${\alpha }_1^{\mathrm{\prime }}(0)={\alpha }_2^{\mathrm{\prime }}(0)$ by [F2], these derivatives are equal for $r=1,2$. Thus $(f\circ {\gamma }_1{)}^{\mathrm{\prime }}(0)=(f\circ {\gamma }_2{)}^{\mathrm{\prime }}(0)$.

Therefore the two curves determine the same velocity derivation.