Whenever $P$ is a poset and $x,y$ are elements of $P$, write $x \vartriangle y$ to mean that $x$ and $y$ have a lower bound. Then $\vartriangle_P$ is always reflexive and symmetric, but it may or may not be transitive.
Transitivity does not hold.
Let $X = \{0,1,2,3\}$. Let $P$ denote the poset of all non-empty subsets of $X$. Then $\vartriangle_P$ is not transitive; since letting $x,y$ and $z$ denote $\{0,1\},\{1,2\}$ and $\{2,3\}$ respectively, we have that $x \vartriangle y$ and $y \vartriangle z,$ but $\neg(x \vartriangle z).$
Transitivity holds.
Let $X$ and $Y$ denote sets, and write $P$ for the poset of all partial functions $X \rightarrow Y$ defined at all but finitely many elements of $X$. Then $\vartriangle_P$ is transitive.
Question. Is there a nice characterization, not mentioning $\vartriangle_P$, of those posets in which $\vartriangle_P$ is transitive?
The connected posets where $\vartriangle_P$ is transitive are exactly those where every two elements have a lower bound. So in general a poset has transitive $\vartriangle_P$ exactly if it is the disjoint (i.e. non-related) union of components where in each component every two elements have a lower bound.
It is difficult to see how one could say that without somehow using the notion of elements having a common lower bound.
An example of an order that has this property without (apparently) being "nice" in any other way would be $\mathbb Z\times\mathbb Z$ with the ordering $$(a,p)\sqsubseteq (b,q) \quad\iff\quad b-a \ge |q-p|$$ Note that this doesn't have greatest lower bounds for all pairs, and there are infinite antichains with no lower bounds at all.