A $\lambda$-supercompactness embedding with critical point $\kappa$ is actually a $\lambda^{{<}\kappa}$-supercompactness embedding meaning that $j(\kappa) > \lambda^{{<}\kappa}$ and $M^{\lambda^{{<}\kappa}} \subseteq M$ so once you verify that $D$ has size at most $\lambda^{{<}\kappa} < j(\kappa)$ in $V$, you do so for $M$ as well.  Therefore $\bigcup D \in j(C)$ by the ${<}j(\kappa)$-directed closure of $j(C)$ in $M$, as you mention.

If $x \in C$, then $|x| < \kappa$ and $x \subseteq \lambda$ so $j(x) = j''x \subseteq j''\lambda$.  The equality holds because given a bijection $f: \alpha \rightarrow x$ for some $\alpha$ below the critical point, $j(f)$ will be a bijection between $j(\alpha) = \alpha$ and $j(x)$.  So every element of $j(x)$ is of the form $j(f)(\beta)$ for some $\beta < \alpha$, but $j(f)(\beta) = j(f(\beta)) \in j''x$ since $\beta$ is also below the critical point.