For homogeneous equilibrium, why are liquids and solids included in the equilibrium constant (when they aren't in heterogeneous equilibria)?
Chemguide is simplified for A-levels and therefore in this case is strictly speaking incorrect.

The equilibrium constant $K$ is defined as a product of activities. I described this in a previous answer here.

The crux of the matter is that the activity of a pure solid or pure liquid is equal to 1, which means that it can be omitted from the expression for $K$ without affecting the value.

In your first reaction

$$\ce{H2O(g) + C(s) &lt;=&gt; H2(g) + CO(g)}$$

the chunks of carbon in the reaction are necessarily pure because they don't mix with the gases.

In the second reaction (yes, I am lazy, please feel free to edit for me)

$$\ce{EtOAc(l) + H2O(l) &lt;=&gt; AcOH(l) + EtOH(l)}$$

none of the liquids are pure, hence their activities deviate from unity.

Just as a final example, in the dissociation of a weak acid

$$\ce{HA(aq) + H2O(l) &lt;=&gt; H3O+(aq) + A-(aq)}$$

water is omitted from the expression for $K_\mathrm{a}$ because water, as the solvent, is in large excess over $\ce{HA}$ and is therefore effectively "pure".