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With the $+---$ signature and $n^2=1$, direct expansion of the two spatial projection tensors givesThis restriction is negative definite on spatial vectors. The positive spatial induced metric is instead $h_{\alpha \beta }=n_{\alpha }{n}_{\beta }-g_{\alpha \beta }=-q_{\alpha \beta }$, whose pullback to a time slice is the $h_{ij}$ used in the line element. The spatial metric sign for a unit timelike normal is important here: the plus sign in the PDF's claimed equality is incompatible with its normal normalization: $g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ is not even transverse to $n^{\alpha }$.

The spatial covariant derivative of a spatial tensor projects every index, including its derivative index. Projection only on the derivative index is sufficient for a scalar but not for a general tensor. Using metric compatibility of the spacetime Levi-Civita connection, we obtainEach term contains a normal contracted with its spatial projection tensor. Consequently $\boxed{D_i{h}_{jk}=0}$, and the negative spatial restriction also satisfies $D_i{q}_{jk}=0$. This proves the requested metric compatibility of the spatial covariant derivative after correcting the source's metric sign. Contracting indices gives the particular expression written in the question.

Using the normalization of the timelike unit normal, the spatial projection tensor satisfiesLikewise $P^{\mu }{}_{\alpha }{n}^{\alpha }=0$ and $P^{\mu }{}_{\alpha }{P}^{\alpha }{}_{\nu }=P^{\mu }{}_{\nu }$. The spacetime form of the induced metric is thereforeIts restriction to the slice is the given three-metric. Metric compatibility of the four-dimensional Levi-Civita connection givesApply the three projectors defining the spatial covariant derivative. In the first summand $P^{\beta }{}_j{n}_{\beta }=0$, and in the second $P^{\alpha }{}_i{n}_{\alpha }=0$. Both summands vanish, soThis proves the metric compatibility of the spatial covariant derivative without treating the spatial projection tensor itself as covariantly constant.