An epimorphism from $S_{4}$ to $S_{3}$ having the kernel isomorphic to Klein four-group

Yes, there is. My favorite way of doing that is the following. There are exactly three ways of partitioning the set $\{1,2,3,4\}$ to two disjoint pairs, namely $$ P_1=\{\{1,2\},\{3,4\}\},\quad P_2=\{\{1,3\},\{2,4\}\},\quad\text{and}\quad P_3=\{\{1,4\},\{2,3\}\}. $$ Now given a permutation $\sigma\in S_4$ it acts naturally on the set $\{P_1,P_2,P_3\}$ of such partitions "elementwise", and thus gives us a permutation $\overline{\sigma}\in Sym(\{P_1,P_2,P_3\})$. This correspondence $\sigma\mapsto \overline{\sigma}$ is (one of) the epimorphism(s) you are looking for.

More details: $\overline{\sigma}$ takes the partition $P_1$ to the partion $\{\{\sigma(1),\sigma(2)\},\{\sigma(3),\sigma(4)\}\}$ and similarly for the others. For example, when $\sigma=(234)$ we get that $$ \begin{aligned} \overline{\sigma}(P_1)&=\{\{1,3\},\{4,2\}\}=P_2,\\ \overline{\sigma}(P_2)&=\{\{1,4\},\{3,2\}\}=P_3,\\ \overline{\sigma}(P_3)&=\{\{1,2\},\{3,4\}\}=P_1,\\ \end{aligned} $$ so the resulting permutation is the 3-cycle $P_1\mapsto P_2\mapsto P_3\mapsto P_1$.

It is tedious but straightforward to check that the mapping $\sigma\mapsto \overline{\sigma}$ is surjective. It is a bit easier to check the all the permutations of the subgroup $N$ leave all the partitions $P_j,j=1,2,3,$ invariant.

Take a tetrahedron. The symmetries form the group $S_4$ on the vertices. Consider the action of these symmetries on the three pairs of opposite edges of the tetrahedron. i.e. Pair 1 - edges 12, 34; Pair 2 edges 13, 24; pair 3 edges 14, 23.

I'll leave you to work out the details. The other platonic solids also give some geometric realisations of other relationships between groups.