Derivations:
I've got the whole thing in notebooks progressing the other way, from the EFE to the ADM,
following Baumgarte & Shapiro, so TODO copy it here.
ADM
$\gamma_{ij,t} - \mathcal{L}_\vec\beta \gamma_{ij} = -2 \alpha K_{ij}$
$K_{ij,t} - \mathcal{L}_\vec\beta K_{ij} = -D_i D_j \alpha + \alpha (R_{ij} + K K_{ij} - 2 K_{ik} {K^k}_j)$
expanding Lie derivatives and projection covariant derivatives:
$\gamma_{ij,t} - \gamma_{ij,k} \beta^k - \gamma_{ik} {\beta^k}_{,j} - \gamma_{kj} {\beta^k}_{,i} = -2 \alpha K_{ij}$
$K_{ij,t} - \beta^k K_{ij,k} - K_{kj} {\beta^k}_{,i} - K_{ik} {\beta^k}_{,j} = -{\perp_i}^a \nabla_a ( {\perp_j}^b \nabla_b \alpha ) + \alpha (R_{ij} + K K_{ij} - 2 K_{ik} {K^k}_j)$
$= -{\perp_i}^a \nabla_a ( \alpha_{,j} - n_j n^b \alpha_{,b} ) + \alpha (R_{ij} + K K_{ij} - 2 K_{ik} {K^k}_j)$
$= -{\perp_i}^a ( \nabla_a \alpha_{,j} - \nabla_a (n_j n^b \alpha_{,b}) ) + \alpha (R_{ij} + K K_{ij} - 2 K_{ik} {K^k}_j)$
$= -{\perp_i}^a ( \nabla_a \alpha_{,j} - \nabla_a n_j n^b \alpha_{,b} - n_j \nabla_a n^b \alpha_{,b} - n_j n^b \nabla_a \alpha_{,b} ) + \alpha (R_{ij} + K K_{ij} - 2 K_{ik} {K^k}_j)$
$= -(\delta_i^a - n_i n^a) ( \alpha_{,ja} - {^4\Gamma^b}_{ja} \alpha_{,b} - (n_{j,a} - {^4\Gamma^c}_{ja} n_c) n^b \alpha_{,b} - n_j ({n^b}_{,a} + {^4\Gamma^b}_{ca} n^c) \alpha_{,b} - n_j n^b (\alpha_{,ba} - {^4\Gamma^c}_{ba} \alpha_{,c}) ) + \alpha (R_{ij} + K K_{ij} - 2 K_{ik} {K^k}_j)$
...is $D_i$ the same as applying the spatial connections?
$K_{ij,t} - \beta^k K_{ij,k} - K_{kj} {\beta^k}_{,i} - K_{ik} {\beta^k}_{,j}$
$= -D_i \alpha_{,j} + \alpha (R_{ij} + K K_{ij} - 2 K_{ik} {K^k}_j)$
$= -\alpha_{,ij} + {\Gamma^k}_{ji} \alpha_{,k} + \alpha (R_{ij} + K K_{ij} - 2 K_{ik} {K^k}_j)$