Prove $D_{24}$ is not isomorphic to $S_4$.

$D_{24}$ and $S_4$ are not isomorphic because $D_{24}$ has an element of order $12$ but $S_4$ doesn't.

To show that two groups are isomorphic, it is not sufficient to find a homomorphism between them, even if they have the same order. For instance, you can always pick the trivial homomorphism, which is never an isomorphism (unless both groups are trivial). For a homomorphism to be an isomorphism, it has to be injective and surjective. If you know beforehand that the two finite groups have the same order, then injectivity is equivalent to surjectivity, so you only have to check one.

That being said, I think you had the following idea: define a homomorphism $\phi:D_{24} \to S_4$ by declaring that $\phi(r) = \sigma$ and $\phi(s) = k$ and extending this to the whole group. To be precise, I will use the generators $\sigma = (1234), k=(12)$ for $S_4$. At a first glance, it seems fine because order considerations show that the relations $r^{12} =1$ and $s^{2} = 1$ are preserved. However, this $\phi$ is not well-defined because the third relation $(rs)^2 = 1$ in $D_{24}$ is not preserved. In other words, if $\phi$ were to be a homomorphism, we should have $1 = \phi(1) = \phi((rs)^2) = (\sigma k)^2$. However, $(\sigma k)^2 = (143)$ is not the identity element in $S_4$. The moral of this story is: if you want to define a homomorphism on the generators of a group defined via a presentation, you have to make sure that all the relations are preserved by your would-be homomorphism.

$D_{24}$ and $S_4$ are not isomorphic, because $S_4$ has trivial center, but $D_{24}$ does not - see this question: Center of dihedral group

Two finite groups (of the same order) being "homomorphic" does not mean they are "isomorphic".