The sum construct on cpo A + B denotes a disjoint union of two cpos with a new, dedicated \bot value:
A + B = \{ (0, a) \mid a \in A \} \cup \{ (1, b) \mid b \in B \} \cup \{ \bot _{A+B} \}
This construct is a cpo under the ordering:
x \sqsubseteq y\;\;\text {iff}\;\;\begin {cases} \text {true} & \text {if}\ x=\bot _{A+B}\\ a \sqsubseteq a' & \text {if}\ x = (0,a)\ \text {and}\ y = (0, a') \\ b \sqsubseteq b' & \text {if}\ x = (1,b)\ \text {and}\ y = (1, b') \\
\text {false} & \text {otherwise} \end {cases}
Intuitively, this says that the ordering on A + B is the same as A and B separately, except that \bot _{A+B} is less than anything.
The only directed sets are those that do not contain a mixture of elements from both A and from B, so the least upper bound operator is essentially the same as the least upper bounds for A and for B. Formally:
\bigsqcup X \;\; = \;\; \begin {cases}
(0,\bigsqcup \{ a \mid (0,a) \in X \}) & \text {if}\ \{ a \mid (0,a) \in X \} \neq \emptyset \\
(1,\bigsqcup \{ b \mid (1,b) \in X \}) & \text {if}\ \{ b \mid (1,b) \in X \} \neq \emptyset \\
\bot _{A+B} & \text {otherwise}\\
\end {cases}