"Atransitive" relations appear very frequently. I'll give you two mathematical examples.
The immediate successor relation on natural numbers is atransitive: $x R y$ holds precisely if $x$ is one less than $y$. If $x$ is one less than $y$ and $y$ is one less than $z$, then $x$ cannot be one less than $z$.
On any bipartite graph, the edge relation is atransitive: if $x R y$ holds, then $x$ belongs to one part, and $y$ to the other part. If $y R z$, then $z$ belongs to the same part as $x$, so there cannot be an edge between them.
You can easily turn this into examples from ordinary life: consider e.g. $xRy$ that holds if $x$ is standing immediately behind $y$ in a queue; or the relation $xRy$ that holds if $x$ and $y$ are playing in opposite teams in a game of football.
We can also find some "reverse-transitive" relations on arbitrarily large sets. For example, the relation that never holds satisfies reverse-transitivity on any set. The relation that always holds also satisfies the condition, as does every symmetric, transitive relation.
Mandating asymmetry, by itself, is not sufficient to characterize rock-paper-scissors: you can have asymmetric, reverse-transitive relations on sets with more than three elements. Consider e.g. the "$x$ loses against $y$" relation of the game rock-paper-scissors-MathSEkarma, where rock, paper, scissors beat each other as usual, while MathSEkarma is completely inert: it doesn't beat and is not beaten by anything. Or rock-metal-paper-scissors, where rock, paper, scissors work as usual, and metal works exactly like rock, winning only against scissors but losing only against paper. These both satisfy reverse-transitivity.
To obtain a unique characterization of rock-paper-scissors, we need the relation $R$ to satisfy at least one more property beyond asymmetry. For example, the condition $xRy \vee yRx \vee x = y$ does the trick (exercise!).
If there are four or more elements,then we can find some $x$ that loses against (i.e. is related to) two different elements $y$,$z$. But then y cannot lose against z, as otherwise by reverse-transitivity we would have that $x$ does not lose against $z$. Similarly, $z$ cannot lose against $y$ either, which violates the condition above.