You can find many such examples among groups acting on trees. Let $T$ be a $k$-regular tree and let $G$ be the subgroup of $\operatorname{Aut}(T)$ of automorphisms acting without inversion. (This is a simple subgroup of $\operatorname{Aut}(T)$ of index $2$). Note that $G$ is a totally disconnected locally compact group w.r.t. the permutation topology. Fix a coloring $\lambda \colon E(T) \to \{ 1, \dots, k \}$ of the edges of the tree. We use the notation $T_v$ for the star around a vertex $v$, i.e., the collection of $k$ edges having $v$ as one of its end points.

Now fix a finite permutation group $F \leq \operatorname{Sym}(k)$. Then we can consider the Burger-Mozes universal group $$ U(F) = \{ g \in G \mid \lambda_{|T_{g.v}} \circ g \circ \lambda_{|T_v}^{-1} \in F \text{ for all } v \in V(T) \} ; $$ intuitively, this is the largest subgroup of $G$ with prescribed local action equal to $F$. Clearly, it is a closed subgroup.

Then $U(F)$ always acts transitively on each of the two partitions of the vertex set $V(T)$, so it is cocompact. Moreover, it is non-discrete as soon as $F$ does not act freely on $\{ 1,\dots,k \}$, and in addition, it is simple if (and only if) $F$ is transitive and generated by its point stabilizers. In particular, it is unimodular in this case.

You can find many such examples among groups acting on trees. Let $T$ be a $k$-regular tree and let $G$ be the subgroup of $\operatorname{Aut}(T)$ of automorphisms stabilizing each of the 2 parts of the bipartition of the vertex set $V(T)$. (This is a simple subgroup of $\operatorname{Aut}(T)$ of index $2$). Note that $G$ is a totally disconnected locally compact group w.r.t. the permutation topology. Fix a coloring $\lambda \colon E(T) \to \{ 1, \dots, k \}$ of the edges of the tree. We use the notation $T_v$ for the star around a vertex $v$, i.e., the collection of $k$ edges having $v$ as one of its end points.

Now fix a finite permutation group $F \leq \operatorname{Sym}(k)$. Then we can consider the Burger-Mozes universal group $$ U(F) = \{ g \in G \mid \lambda_{|T_{g.v}} \circ g \circ \lambda_{|T_v}^{-1} \in F \text{ for all } v \in V(T) \} ; $$ intuitively, this is the largest subgroup of $G$ with prescribed local action equal to $F$. Clearly, it is a closed subgroup.

Then $U(F)$ always acts transitively on each of the two partitions of the vertex set $V(T)$, so it is cocompact. Moreover, it is non-discrete as soon as $F$ does not act freely on $\{ 1,\dots,k \}$, and in addition, it is simple if (and only if) $F$ is transitive and generated by its point stabilizers. In particular, it is unimodular in this case.