In a Riemannian manifold $\mathcal{S}$ with metric $\boldsymbol{g}$, given a chart $\{x^a\}$, it is fairly easy to prove that the divergence of a vector field $\boldsymbol{w} : \mathcal{S} \to T\mathcal{S}$ is given by $$ \mathrm{div}\boldsymbol{w} = w^a{}_{|a} = \frac{1}{\sqrt{\det[g_{mn}]}} \frac{\partial}{\partial x^a} \left[ \sqrt{\det[g_{mn}]} \, w^a \right], $$ where $[g_{mn}]$ is the matrix representation of $\boldsymbol{g}$ in the chart $\{x^a\}$, and the vertical bar in ${}_{|a}$ denotes the covariant derivative with respect to the Levi-Civita connection induced by $\boldsymbol{g}$.
I am wondering whether there is an equivalent expression for the divergence of a higher-order tensor, say, a second-order tensor $\boldsymbol{\sigma} : \mathcal{S} \to T\mathcal{S} \otimes T\mathcal{S}$, in which case the component form is $(\mathrm{div}\boldsymbol{\sigma})^a = \sigma^{ab}{}_{|b}$, with the divergence taken, e.g., on the last index.
PS: In case you are curious, yes, I am looking for an expression of the divergence of the (fully contravariant) Cauchy stress.
There is indeed such expression, you have written it yourself: $\sigma^{ab}|_{b}$. Now it is a matter of writing the Christoffel symbols in the form you want, on the one hand the covariant derivative acts on both indices $a,b$ so the answer is:
$$ \sigma^{ab}|_{b} = \frac{1}{\sqrt{\det g}}\partial_b (\sqrt{\det g} \sigma^{ab}) + \Gamma^{a}_{bc}\sigma^{cb} $$ but I doubt that the second term could be made into a total derivative because it invloves all the components of the Christoffel symbol.