How to tell/control whether sodium ethoxide will react by a substitution or elimination pathway?
The above reactions belong to Williamson Synthesis of Ethers, probably the best of the alternatives for prepartion of ethers!

The Williamson Ether synthesis reactions follow SN2 mechanism. Since the SN2 mechanism proceeds through a single step where the nucleophile performs a “backside attack” on the alkyl halide, the only thing stopping this, is steric hindrance.

Methyl and primary alkyl halides are excellent substrates for this synthesis, as they provide no/very less hindrance for the approaching nucleophile.

Since alkoxides are strong bases, competition with elimination (E2) becomes a concern once the alkyl halide becomes more sterically hindered. This is exactly why, the first reaction that you’ve mentioned, proceeds by SN2 but in the second both elimination(major) and substitution (minor) takes place.

Also note that, when tertiary alkyl halides are used, there is no hope for substitution because of the bulky groups present around carbon. Hence, only elimination takes place in this case.



So the reactions are,


With primary alkyl halide,
$$ \ce{CH3CH2CH2CH2Br + C2H5ONa -&gt; CH3CH2CH2CH2-O-CH2CH3}$$
With secondary alkyl halide,
$$\ce{CH3CH2CHBrCH3 + C2H5ONa -&gt; CH3CH2CH=CH2(major)}$$




Moving onto conditions to favour a particular type of reaction:


One way to attempt to get the SN2 to be favoured over elimination is to use a polar aprotic solvent (such as acetonitrile or DMSO) that will increase the nucleophilicity of the alkoxide. 
Increasing the temperature favours elimination over substitution.


Source for the above conditions: chem.ucalgary




  I believe sodium ethoxide being bulky should always do elimination


In most cases, rather than the incoming nucleophile, the hindrance in the the alkyl halide plays a more important role.



The rule #3 mentioned in the link you stated clearly says that the elimination takes place for a tertiary carbon in presence of a strong base like $\ce{CH3CH2-O-Na}$.