kcat in uncompetitive enzyme inhibition
Short answer
It all depends on how you look at $k_\text{cat}$ and enzyme concentrations

Long answer
The effect of an uncompetitive inhibitor on $k_\text{cat}$ can be interpreted in two ways.
Interpretation 1:
This is followed in the lecture series mentioned in the question.
See the video https://youtu.be/rNBEUGYu034 from 7:37 to 11:01 minutes. The instructor says:
... the turnover number basically describes the ability of that active site to actually transform the substrate molecules into the product molecules per unit time. Now because when the inhibitor is not bound to that particular enzyme-substrate complex, the active site's ability or efficiency to change that substrate to the product doesn't actually change, we see that the $k_\text{cat}$ value in uncompetitive inhibition also doesn't actually change...
Mathematically, $k_\text{cat}=\frac{V_\text{max}}{[\ce{E}]}$ (where $[\ce{E}]$ is the concentration of enzyme, excluding that which has been inactivated by inhibitor). Now, uncompetitive inhibition reduces both $V_\text{max}$ and $[\ce{E}]$ by the same factor. Thus $k_\text{cat}$ remains unchanged.
Interpretation 2:
Some people do things differently. They regard $k_\text{cat}$ as $\frac{V_\text{max}}{[\ce{E}]}$, where $[\ce{E}]$ is the total concentration of enzyme in the mixture (including inactivated forms). In this treatment, the uncompetitive inhibitor reduces $V_\text{max}$ but $[\ce{E}]$ stays the same. Thus $k_\text{cat}$ is reduced.
Qualitatively, the enzyme in the mixture is now a combination of active and inactivated forms, and thus has lower &quot;efficiency&quot; overall, i.e. lower $k_\text{cat}$.