Property of a totally ordered Field

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One of the properties of a totally ordered Field $K$ is

For all $ x , y , u , v \in ( K , > )$, then: $$x < y \wedge u < v \rightarrow x + u < y + v$$

I'm not sure what the above property is true because it is $y \wedge u $ that is greater than $x$ and smaller than $y$ but not $y$ and $u$ itself. How do we come to the implication? Are $ x, y, u, v$ sets within $K$ ?because the definition does not specify their algebraic structure.

Last but not least, since $y$ and $u$ are not specified, can this be true? $$x < y \wedge u < v \rightarrow x + v < y + u$$

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I think you are reading the statement wrong. It says:

$$(x < y) \wedge (u < v) \rightarrow x + u < y + v$$