We say that and ideal $\mathcal I$, in a boolean algebra $\mathcal B$ $\kappa$-dense if $\mathcal B/\mathcal I = \{[x]\mid x\in\mathcal B \wedge [x] = \{y\mid x\triangle y \in\mathcal I\}\}$ has a dense subset of power at most $\kappa$. And $\mathcal I$ is $\lambda$-saturated whenever $\{[x_\xi]\mid \xi<\lambda\}\subset\mathcal B/\mathcal I$ is a collection of disjoint sets, that is $[x_\xi]\cap[x_\zeta]=[0]$, then there is $\zeta<\lambda$ with $[x_\zeta]=[0]$. This is equivalent to take an almost disjoint family modulo $\mathcal I$, i.e. $x_\xi\cap x_\zeta\in\mathcal I$, and find $\xi<\lambda$ with $x_\xi\in\mathcal I$.
I need to show that $\kappa$-denseness implies $\kappa^+$-saturation. Here is my try, but i am stuck even a little lost
Take $D\subset\mathcal P(X)/\mathcal I$ dense with $|D|\leq\kappa$, $D=\{[d_\xi]\mid\xi<\kappa\}$. Let $\{ x_\xi\mid \xi<\kappa^+\}$ be an almost disjoint family, modulo $\mathcal I$. Assume that for $\xi<\kappa^+$, $x_\xi\notin \mathcal I$ and let $W=\{[x_\xi]\mid \xi<\kappa^+\}$, but $[x_\xi]=\{y\subset X\mid x_\xi\triangle y \in \mathcal I\}$ take $\xi,\zeta<\kappa^+$, we state that $x_\xi\triangle x_\zeta\notin \mathcal I$ since if this were not the case, as $x_\xi\triangle x_\zeta=(x_\xi\setminus x_\zeta)\cup(x_\zeta\setminus x_\xi)$,
$$ (x_\xi\setminus x_\zeta),(x_\zeta\setminus x_\xi), x_\xi\cap x_\zeta \in\mathcal I; $$ $$ (x_\xi\setminus x_\zeta)\cup ( x_\xi\cap x_\zeta) = (x_\xi\cap (X\setminus x_\zeta))\cup ( x_\xi\cap x_\zeta)= $$ $$ (x_\xi\cup (x_\xi\cap (X\setminus x_\zeta)))\cap (x_\zeta\cup (x_\xi\cap (X\setminus x_\zeta))); $$ $$ \text{but } x_\xi\cup (x_\xi\cap (X\setminus x_\zeta)) = x_\xi, $$ $$ \text{and }x_\zeta\cup (x_\xi\cap (X\setminus x_\zeta)) = (x_\zeta\cup x_\xi)\cap (x_\zeta \cup (X\setminus x_\zeta) = (x_\zeta\cup x_\xi); $$ $$ \text{therefore } (x_\xi\setminus x_\zeta)\cup ( x_\xi\cap x_\zeta) = x_\xi\cup(x_\zeta\cap x_\xi) = x_\xi \in\mathcal I. $$
a contradiction, similarly we obtain $(x_\zeta\setminus x_\xi)\cup ( x_\xi\cap x_\zeta)=x_\zeta\in\mathcal I$. Thus $\{x_\xi\}_{\xi<\kappa^+}$ generates $\kappa^+$ distinct classes. For $\xi<\kappa^+$ let $[d_{\alpha_\xi}]\in D$ be such that $[d_{\alpha_\xi}]\subset [x_\xi]$; thus there is $Y\subset\kappa^+$ with $|Y|=\kappa^+$ and for $\xi,\zeta\in Y$, $[d_{\alpha_\xi}]=[d_{\alpha_\zeta}]$, we denote this unique element as $[d_Y]$. Therefore $[d_Y]\subset [x_\xi],\ \xi\in Y$.
It seems to me that you essentially have the proof at the end of your question, but it's preceded by a lot of extraneous stuff. Work in the Boolean algebra $\mathcal B/\mathcal I$. Assume it has a dense subset $D$ of cardinality $\kappa$, and suppose we're given $\kappa^+$ elements $x_\xi$. Each $x_\xi$ is above some $y_\xi\in D$ because $D$ is dense. And because $|D|=\kappa$, these $\kappa^+$ elements $y_\xi$ can't all be distinct. So we have some $x_\xi$ and $x_\eta$ that are both above the same $y_\xi=y_\eta\in D$ and therefore cannot be disjoint (in $\mathcal B/\mathcal I$).