Suppose you have two fluid bodies, one described by a set of vectors $V$, and a perturbation of $V$ given by $V+\Delta V$. Suppose that the two regions are related by the transformation $\mathbf x-\xi(\mathbf x,t)=\mathbf x'$ where $\mathbf x\in V+\Delta V$ and $\mathbf x'\in V$. Given some scalar function $Q(\mathbf x,t)$, consider the integral $\int_{V+\Delta V}Q(\mathbf x,t)d\mathbf x$. The change of variables described above may be used to express the integral over the unperturbed region $V$, as $$\int_{V}Q(\mathbf x'+\xi,t)(1+\operatorname {div}\xi )d\mathbf x'$$ where $1+\operatorname {div} \xi$ is the jacobian of the transformation.
From a conceptual perspective, it makes sense to have the jacobian given by $1+\operatorname{div}\xi$, since we're measuring the infinitesimal change in volume by taking the original volume increment (corresponding to the 1 in the jacobian), and adjusting according to how much of the fluid flowed into or out of it (the div in the jacobian).
But I can't figure out to how actually calculate this using the change of variables shown here - I keep getting values twisted around in my calculations. Any help is appreciated.
I believe you are using the approximation $\det(1+A) = 1 + \operatorname{tr} A$ which is valid as long as A is small, which is true for short time. One way of obtaining this approximation is by computing the Fréchet derivative of $\det$ as $d\det (B) A = \operatorname{tr}(\operatorname{cof}(B)^T A)$, and then computing the first order Taylor expansion around $B = I$,
\begin{align} \det(I+A) &= \det I +\operatorname{tr}(\operatorname{cof}(I)^T A) + o(A) \\ &= 1 + \operatorname{tr}A + o(A) \end{align}
I believe $\xi$ should depend on the initial position of the fluid particle? So the relation should be $\mathbf{x} = \mathbf{x}'+\xi(\mathbf{x}',t)$, and we see that the transformation gives us $$\text{d}\mathbf{x} = \text{d}\mathbf{x}'\det[I + J\xi(\mathbf{x}',t)] $$
Plugging in our approximation for $\xi\ll 1$ gives us the required relationship, as $\operatorname{tr} J\xi = \partial_i\xi_i = ∇\cdot\xi $.