Their convention for the partition function is explained in the second paragraph before the equation (B.7) of the paper by Shiraz et al. It is
$$z(x) = \sum_{operators} x^{\Delta_{operator}}$$
Note that this is just a different way of writing the usual $\mbox{Tr }\exp(-\beta H)$ if you identify $\exp(-\beta)\equiv x$ and $\Delta\equiv H$. Yes, the dimension is the same thing as the Hamiltonian (of the radial quantization) and it is often helpful to avoid exponentials and write powers of $x$ only, so therefore the exponential redefinition of $\beta$ vs $x$. The trace is the summation over the basis.
They're calculating the partition function of a whole theory, not the $\partial$ operator itself. So the partition function is the sum over operators, as described above. In this simple case, the operators are $\partial_i \partial_j \dots \phi$, i.e. arbitrary derivatives of $\phi$ by $d$ different partial derivative symbols.
The derivatives with respect to different directions commute with each other and are completely independent. So imagine $d=1$ for a while, only one direction. In that case, you have operators
$$\phi, \partial \phi, \partial^2 \phi, \dots$$
and their dimensions are
$$\Delta=0,1,2,\dots$$
plus the dimension of $\phi$ if it were nonzero. The partition sum is the sum of $x^\Delta$ over these operators which means
$$1+x+x^2+\dots = \frac{1}{1-x}.$$
The sum is obtained as geometric series. Note that the coefficients of the Taylor expansion are simply equal to one: there is no source where you could have gotten something else.
Now, the operators in the $d$-dimensional space may be obtained by acting with some derivatives in the 1st direction; some in 2nd, and so forth, on $\phi$. So the space of operators is a tensor product of spaces from each of the $d$ directions, and the partition sum is therefore the product of the partition sums from the individual directions, i.e. the $d$-th power of $1/(1-x)$.
There are other factors multiplying the total partition function but you haven't asked about it, and I can't explain every detail in a 50-page paper you haven't asked about. But yes, the other paper you mentioned almost certainly uses the same basic insight about the geometric series.
