I would like to get an intuitive idea as to why the definitions of order-dense and locally order-dense Riesz subspaces were chosen in the following way:
A Riesz subspace $F$ of $E$ is order-dense if for every $x \in E^{+},$ $$x =\sup \{y:~0 \leq y \leq x, ~ y \in F \}.$$
$F$ is locally order-dense if it is order-dense in its solid hull i.e., if $$0 \leq x \leq x_{0} \in F \implies x = \sup\{ y:~ y\in F, 0 \leq y \leq x \}.$$
Thanks for any assistance.
A Riesz subspace can be characterised as being regular, $\sigma -$regular or majorizing depending on how it relates to the Riesz space $E$ (either through embeddings of the spaces or elements within each space).
One of its important features is that if our Riesz subspace $F$ is regular (ie, embedding the subspace into $E$ preserves supremum and infinum), then this gives us conditions for the convergence of a sequence $\{u_a\}$ in $E$ and subspaces.
The definition you have given for Riesz subspaces can be rewritten as "For $x\in E^+,~\exists$ a net $\{ u_a \}\subseteq F$ such that $0\leq u_a$ converges pointwise (increasingly) to $u$ in $E$".
Let $A^d$ is a nonempty subset of $E$ and $A^{dd}$ is its disjoint complement. Combining the above fact along with the fact that every ideal $I$ of $E$ is order dense in $A^{dd}$, we can conclude that $A=A^{dd}$, ie, the disjoint complement of the disjoint complement of $A$ is $A$.