Does this work?

To prove $D(\alpha)$ is $\kappa$-compact for all $\alpha$ in $A$, assume otherwise, that there exists some counterexample. Then, by the fact $A$ is well-ordered, there is a minimal counterexample (i.e., there is a minimal element $\alpha$ in the set of $\gamma \in A$ such that $D(\gamma)$ is not $\kappa$-compact). This means $D_\beta$ is $\kappa$-compact for all $\beta \lt \alpha$. Since $\{\beta: \beta \lt \alpha\}$ has cardinality less than $\kappa$, we have that

$$colim_{\beta: \beta \lt \alpha} D(\beta)$$

is $\kappa$-compact. Now, given a diagram of the form

$$V_\alpha \leftarrow U_\alpha \to colim_{\beta: \beta \lt \alpha} D(\beta)$$

in the category of $\kappa$-compact objects, its pushout is also $\kappa$-compact. But the hypothesis is that $D(\alpha)$ is the pushout for some such diagram, so $D(\alpha)$ is $\kappa$-compact, and we have reached a contradiction.

So $D(\alpha)$ is $\kappa$-small for all $\alpha \in A$. It follows that $colim_{\beta \in B} D(\beta)$ is $\kappa$-compact for any subposet $B \subseteq A$ whenever this is a $\kappa$-small colimit. (The restriction to downward-closed $B$ is not much loss of generality, because if $B \subseteq A$ is full, then the colimit over such a $B$ is isomorphic to the colimit over its downward closure, since $B$ is cofinal in its downward closure.)

Does this work?

To prove $D(\alpha)$ is $\kappa$-compact for all $\alpha$ in $A$, assume otherwise, that there exists some counterexample. Then, by the fact $A$ is well-ordered, there is a minimal counterexample (i.e., there is a minimal element $\alpha$ in the set of $\gamma \in A$ such that $D(\gamma)$ is not $\kappa$-compact). This means $D_\beta$ is $\kappa$-compact for all $\beta \lt \alpha$. Since $\{\beta: \beta \lt \alpha\}$ has cardinality less than $\kappa$, we have that

$$colim_{\beta: \beta \lt \alpha} D(\beta)$$

is $\kappa$-compact. Now, given a diagram of the form

$$V_\alpha \leftarrow U_\alpha \to colim_{\{\beta: \beta \lt \alpha\}} D(\beta)$$

in the category of $\kappa$-compact objects, its pushout is also $\kappa$-compact. But the hypothesis is that $D(\alpha)$ is the pushout for some such diagram, so $D(\alpha)$ is $\kappa$-compact, and we have reached a contradiction.

So $D(\alpha)$ is $\kappa$-small for all $\alpha \in A$. It follows that $colim_{\beta \in B} D(\beta)$ is $\kappa$-compact for any subposet $B \subseteq A$ whenever this is a $\kappa$-small colimit. (The restriction to downward-closed $B$ is not much loss of generality, because if $B \subseteq A$ is full, then the colimit over such a $B$ is isomorphic to the colimit over its downward closure, since $B$ is cofinal in its downward closure.)