By definition, $(x, z)\in S\circ R$ if there exists a $y$ such that $(x,y)\in R$ and $(y,z)\in S$. This is easy to visualize if $R=S$ and the relation $R$ is depicted as a directed graph (possibly with loops). That is, if $R$ is a binary relation on a set $A$, then elements of $A$ are drawn as vertices (or nodes, or points), and there is an arrow from vertex $x$ to vertex $y$ iff $(x,y)\in R$. One can take walks in such a graph by following arrows. Then $(x,z)\in R\circ R=R^2$ if one can make two steps following arrows of $R$: one from $x$ to some $y$ and the other from $y$ to $z$.

Applied to the case when $R$ is $<$, this means that $(x,z)\in{<}\circ{<}$ if there exists some $y$ such that $x<y<z$. How does this compare with simply $x<z$ when the underlying set is $\Bbb Z$? Which statement implies the other?