On this presentation, the last slide is titled "Non Separable Utility", and the preferences given are
$$ \frac{\left(c^\gamma (1-n)^{1-\gamma}\right)^{1-\sigma}}{1-\sigma}$$
However, I can log-transform them as
$$ \log\left[ c^{\gamma(1-\sigma)} (1-n)^{(1-\gamma)(1-\sigma)} (1-\sigma)^{-1}\right]$$
which are clearly separable. Did I miss something?
You are right, these preferences are separable. A way to see it is to notice that \begin{equation*} U(c,n) \geq U(c',n) \Rightarrow U(c,n') \geq U(c',n') \end{equation*} for any $n,n'$, and \begin{equation*} U(c,n) \geq U(c,n') \Rightarrow U(c',n) \geq U(c',n') \end{equation*} for any $c,c'$. They are even additively separable, as your log-transformation shows.
Looking two slides before, the author provides another utility function, under "Separable Utility" this time:
$$U(c,n) = \ln(c) - a\frac {n^{1+1/v}}{1+1/v}$$
Comparing the two I believe we can safely reach the conclusion that the author had implicitly in mind "additive separability", i.e. a functional form where second partials are zero, irrespective of whether a transformation can separate the arguments of the function. At least in some corners, when one needs the logarithmic transformation to induce separability, one talks about "multiplicatively separable" preferences.
