How to balance the reaction between ethanol and chlorine using oxidation number method?
The key  is to remembering that  $n \cdot m = m \cdot n$.
It seems obvious, but you would not believe how much struggles people suffer when trying to apply this trivial equation to chemical reactions.
In this special case, The total change of oxidation numbers is zero, otherwise, it would mean we have lost some electrons somewhere.
$$\sum{n_i \cdot \Delta ON_i } = 0$$.
The stoichiometric coefficients are in the reciprocal ratio to ( absolute value of ) the oxidation number change ratio.
So the sum of the oxidation number change of carbons in 1 ethanol molecule is $( (+2 - -3) + ( +2 - -1) ) = +5 + +3 = +8$
The sum of the oxidation number change of chlorine in 1 chlorine molecule is $2 \cdot (-1 - 0) = -2$
We have then:
$$n_{\ce{C2H5OH}} \cdot 8 + n_{\ce{Cl2}} \cdot -2 = 0$$
$$n_{\ce{Cl2}} = 4 \cdot  n_{\ce{C2H5OH}} $$
Note that you are missing negative ions of the left side for charge balance, also overall mass balance is missing.

Enumeration of stoichiometric coefficients have also a general, analytical mathematical solution, but personally, I never used it, as it would take much longer time than just enumerating it on the fly by the chemist's heart:

You are solving a set of linear algebraic equations for multiple variables.
Each variable is a stoichiometric coefficient for a reagent or a product.
Each equation represents some kind of balance.
1 equation for the atom count balance for each of the elements
1 equation for the charge balance, if applies.
In most cases there will remain 1 degree of freedom, so 1 stoichiometric coefficient must be chosen and some others will depend on the choice.
E.g. there can be $\ce{n H2 + n F2 -&gt; 2n HF}$, so we choose $\ce{n=1}$ for $\ce{H2}$  and rest will follow.