Geodesic deviation in flat space Suppose that $x^\mu(t,s)$ represents a family of curves.
Let $v^\mu$ represents the the tangent vector to a curve $x^\mu(t,s_0)$ with $s_0$ fixed that is $v^{\mu}=\partial x^{\mu} / \partial t$  and deviation vector is given by $\xi^{\alpha}=\partial x^{\alpha} / \partial s$.
In the usual derivation of the geodesic deviation equation it is showed that $\xi$ is lie transposed through $v$ that is $$L_v\xi=0$$ where $L$ stands for the lie derivative.
To make the discussion more clear let us suppose we are in flat spacetime with  usual Cartesian coordinates $x^0=t,x^1,x^2,x^3$.
We choose  $x^\mu(t,s)=(t,st,0,0)$ and so $v^\mu=(1,s,0,0)$ and $\xi^{\alpha}=(0,t,0,0)$
we have $$L_v\xi=\left[\frac{\partial}{\partial t}+s\frac{\partial}{\partial x^{1}},t\frac{\partial}{\partial x^{1}} \right]=\frac{\partial}{\partial x^1}-\frac{\partial s}{\partial x^1}\frac{\partial}{\partial x^1}$$
So we choose the parameter $s$ for example to $2x^1$ we would have $L_v\xi \ne0$
Isn't this a contradiction?

Away from $t=0$ you can think of $x$ as a regular parametrization of a surface in your 4D space. The vector fields you call $v$ and $\xi$ are just the coordinate vector fields usually called $x_t$ and $x_s$. Their Lie bracket vanishes because $x_{st}=x_{ts}$.
