My textbook refers to $\ce{CuO}$ as the oxidising agent. Is this because the $\ce{Cu}$ in $\ce{CuO}$ is oxidised while $\ce{O}$ is neither oxidised or reduced, so overall it is an oxidising agent?
No. $\ce{CuO}$ is the oxidizing agent (the thing that causes something else to be oxidized) because the $\ce{Cu}$ is reduced (gains electrons, going from oxidation state $+2$ to $0$) in the course of oxidizing the nitrogen of $\ce{NH3}$, which loses electrons and goes from oxidation state $-3$ to $0$.
Conversely, $\ce{NH3}$ is the reducing agent here, because it causes something else to be reduced (here, the $\ce{Cu}$) and loses electrons in the process.
What happens for a compound $\ce{AB}$ where $\ce A$ gets oxidised and $\ce B$ gets reduced?
If only the species $\ce{AB}$ is involved, or in the case where two of the same species $\ce A$ react to form both oxidized and reduced products:
$$
\ce{2A -> Ox + Red}
$$
the reaction is called disproportionation.
Do I call the compound $\ce{AB}$ a reducing agent if the net change in oxidation number (between the two elements $\ce A$ and $\ce B$) is negative?
Strictly in the case of a $1\!:\!1$ compound of $\ce{AB}$ on a mole basis, if the net change in oxidation number is negative, then you would call $\ce{AB}$ an oxidizing agent, because it pulled electrons away from something else (it oxidized that other thing) in order to achieve the net negative change in oxidation number.
For $\ce{AB}$ compounds not in a $1\!:\!1$ molar ratio, a universal principle can't be stated $-$ it depends on the particulars of the mole ratio and the individual changes in oxidation number.
