Proof.
Part ([00KJ]) is a restatement of [Lur09, Cor. 5.2.8.18]. To prove part ([00KK]), assume that \(\mathcal C\) is presentable and let \(S\) be a set of morphisms in \(\mathcal C\) that generates \(\mathcal L\). We note first that \(\mathrm{Fun}(\mathcal I,\mathcal C)\) is presentable by [Lur09, Prop. 5.5.3.6]. Now, for each functor \({\sf pt}\xrightarrow{i} \mathcal I\) (selecting an object of \(\mathcal I\)) we obtain an adjunction 
It follows that \(\mathcal R^\mathcal I\) is precisely the right orthogonal to the (small) space of morphisms \[S'
\coloneqq
\bigsqcup_{i \in \iota_0 \mathcal I}
\bigsqcup_{f \in S}
i_!(f)\] in \(\mathrm{Fun}(\mathcal I,\mathcal C)\). From here, Proposition B.1.14 implies that \((\mathcal L^\mathcal I,\mathcal R^\mathcal I)\) is generated by \(S'\) (and in particular that \(\mathcal L^\mathcal I\) is the smallest saturated class of morphisms containing \(S'\)).
For part ([00KL]), given an n-ary operation \((X_1, \ldots, X_n) \rightarrow X\) in \(\mathcal O\), the induced functor \(\mathrm{Fun}(I_{X_1}, \mathcal C_{X_1}) \times \cdots \times \mathrm{Fun}(I_{X_n}, \mathcal C_{X_n}) \rightarrow\mathrm{Fun}(I_X, \mathcal C_X)\) is computed as the left Kan extension of \(I_{X_1} \times \cdots \times I_{X_n} \rightarrow\mathcal C_{X_1} \times \cdots \times \mathcal C_{X_n} \rightarrow\mathcal C_X\) against \(I_{X_1}\times \cdots \times I_{X_n} \rightarrow I_X\). As the left class of a factorization system, \(\mathcal L\) is closed under colimits in \(\mathcal C\). Therefore, the image of natural transformations \(f_i \in \mathrm{Fun}(I_{X_i}, \mathcal C_{X_i})\) which are componentwise in \(\mathcal L\) will again be componentwise in \(\mathcal L\). ◻