The vector gradient, $\mathbb{L}$, is defined as
$$ (\mathbb{L} W)^{ij} \equiv \nabla^{i} W^{j} + \nabla^{j} W^{i} - \frac{2}{3} g^{ij} \nabla_{k} W^{k} \,, $$
where $\nabla_{i}$ is the covariant derivative compatible with the metric $g_{ij}$.
Now, the vector Laplacian, $\Delta_{\mathbb L}$, is define as:
$$ (\Delta_{\mathbb L} W)^{i} \equiv \nabla_{j} (\mathbb{L} W)^{ij} \,. $$
When I expand the above definition, I get:
\begin{align} (\Delta_{\mathbb L} W)^{i} &\equiv \nabla_{j} (\mathbb{L} W)^{ij} \\ &= \nabla_{j} \left( \nabla^{i} W^{j} + \nabla^{j} W^{i} - \frac{2}{3} g^{ij} \nabla_{k} W^{k} \right) \\ &= \nabla_{j} (\nabla^{i} W^{j}) + (\nabla_{j} \nabla^{j}) W^{i} - \frac{2}{3} (g^{ij} \nabla_{j}) \nabla_{k} W^{k} \\ &= \nabla^{i} (\nabla_{j} W^{j}) + (\nabla_{j} \nabla^{j}) W^{i} - \frac{2}{3} \nabla^{i}( \nabla_{k} W^{k}) \\ &= \nabla^{2} W^{i} + \frac{1}{3} \nabla^{i} (\nabla_{j} W^{j}) \,, \end{align}
where $\nabla^{2} \equiv \nabla_{j} \nabla^{j}$.
However, the correct expression should be:
$$ \nabla^{2} W^{i} + \frac{1}{3} \nabla^{i} (\nabla_{j} W^{j}) + R^{i}_{j} W^{j} \,, $$
where $R^{i}_{j}$ is the Ricci tensor.
What am I missing here?
The covariant derivatives $\nabla_j$ and $\nabla^i$ do not commute. Interchanging them gives rise to the curvature term. In general we have
$$\nabla_j\nabla_iW_k - \nabla_i\nabla_jW_k = W_lR^l_{\ kij}$$
with the curvature tensor $R^l_{\ kij}$. Contracting $j$ and $k$ yields
$$\nabla_j\nabla_iW^j - \nabla_i\nabla_jW^j = W_lR^{lj}_{\ \ \ ij}$$
and hence
$$\nabla_j\nabla^iW^j = \nabla^i\nabla_jW^j + R^i_{\ j}W^j$$
from which the correct result follows.