$$L= \underbrace {1111\dots 111}_{2022}$$\begin{align*} 9L^2+2L &= L(9L+2) \\ &= \underbrace {1111\dots 111}_{2022} \cdot (\underbrace {9999\dots 999}_{2022} +2) \\ &= \underbrace {1111\dots 111}_{2022} \cdot (1\underbrace {0000\dots 000}_{2021} 1) \\ &= \underbrace {1111\dots 111}_{4044} \end{align*}The sum of the digits is $\boxed {4044}$.

We can generalize this. Notice that $$L_{k}= \frac{10^{k}-1}{9}\implies S(9\cdot(L_{k})^2+2\cdot (L_{k})).$$So, we have: $S(9\cdot(\frac{10^{k}-1}{9})^2+2\cdot (\frac{10^{k}-1}{9})\implies S( \cancel{9}\cdot \frac{(10^{k}-1)^2}{\cancelto{9}{81}}+2\cdot\frac{10^{k}-1}{3}\cdot\frac{1}{3}+\frac{1}{9}-\frac{1}{9}$ $$=\underbrace{(\frac{10^{k}-1}{3})^2+2\cdot\frac{10^{k}-1}{3}\cdot\frac{1}{3}+\frac{1}{9}}_{(\frac{10^{k}-1}{3}+\frac{1}{3})^2}-\frac{1}{9}$$ Finally, we'll have $S(9(L_{k})^2+2L_{k})= S((\frac{10^{k}\cancel{-1}\cancel{+1}}{3})^2)-\frac{1}{9})= S(\frac{(10^{k})^2}{9}-\frac{1}{9})= S(\frac{10^{2k}-1}{9}).$ Notice that $S(\frac{10^{2k}-1}{9})= S(L_{2k})$. So, for $k= 2022$, we have $S(L_{4044})= S(\underbrace{1111...1111}_{4044})= \underbrace{1+1+1+...+1+1}_{4044}= 4044.$ Therefore, $S(9(L_{2022})^2+2(L_{2022}))= 4044$.