[OP] If i'm using this formula to calculate the extent of reaction is it then dependent on how i write the stoichometry?
No. The formula is general and, for the $j$th species in a chemical reaction, the amount of $j$ is
$$ n_j = n_{j,0} + \nu_j \xi \tag{1} $$
where $n_{j,0}$ is the initial amount of species $j$.
[OP] Which to me this seems to result in a different value for the extent of reaction according to this formula.
Yes. But the results will be the same. I will illustrate with the case you have given, considering that: (1) there is no carbon dioxide initially, and (2) carbon monoxide is the limiting reactant.
\begin{align}
\ce{2CO(g) + O2(g) -> 2CO2(g)} \quad & \quad \ce{CO(g) + 1/2O2(g) -> CO2(g)}
\tag{R1/R2}
\end{align}
By Eq. (1) and making a difference between $\xi_1$ for $\text{R1}$ and $\xi_2$ for $\text{R2}$
\begin{align}
n_\ce{CO} = n_\ce{CO,0} - 2\xi_1 \quad & \quad
n_\ce{CO} = n_\ce{CO,0} - \xi_2 \tag{2,3} \\
n_\ce{O2} = n_\ce{O2,0} - \xi_1 \quad & \quad
n_\ce{O2} = n_\ce{O2,0} - \frac{1}{2}\xi_2 \tag{4,5} \\
n_\ce{CO2} = 2\xi_1 \quad & \quad n_\ce{CO2} = \xi_2 \tag{6,7} \\
\end{align}
Consider as an example where the limiting reactant is depleted, i.e., $n_\ce{CO} = 0$. By Eqs. (2) and (3) we get that the final extents of reaction are
$$ \xi_{1,\text{max}} = \frac{n_\ce{CO,0}}{2} \quad
\xi_{2,\text{max}} = n_\ce{CO,0} \tag{8,9} $$
But when you go to evaluate the final amounts in Eqs. (4-7) we get the same result
\begin{align}
n_\ce{O2} = n_\ce{O2,0} - \frac{n_\ce{CO,0}}{2} \quad & \quad
n_\ce{O2} = n_\ce{O2,0} - \frac{n_\ce{CO,0}}{2} \tag{10,11} \\
n_\ce{CO2} = n_\ce{CO,0} \quad & \quad n_\ce{CO2} = n_\ce{CO,0} \tag{12,13} \\
\end{align}
Thus, as we can see we are at the same point, however both extents have different bounds
$$ 0 \leq \xi_1 \leq \frac{n_\ce{CO,0}}{2} \quad
0 \leq \xi_2 \leq n_\ce{CO,0} \tag{14,15} $$
if the reactions are irreversible. In my humble opinion, as a good practice, always have a stoichiometric number equal to $1$ for the limiting reactant. It keeps tracks of mistakes.