Question About the Substitution of $N$ for $x$ in a $\lambda$-term, as Defined in "Lectures on the Curry-Howard Isomorphism" (1998)

Both definition 1.1.22 and example 1.2.4 (i) are correct, except for the typo you have already noticed in definition 1.1.22 (they write $M\{x:=N\}$ instead of $M[x:=N]$).

In example 1.2.4 (i), the $\lambda$-terms $(\lambda z.z)\lambda y.y$ and $(\lambda z.z)\lambda z.z$ are identical (as $\lambda$-terms). Indeed, a $\lambda$-term is a class of equivalence of pre-terms modulo $\alpha$-equivalence, which roughly means that $\lambda$-terms identify pre-terms that differ only for renaming of bound variables. Now, $\lambda z.z =_\alpha \lambda y.y$ (as pre-terms), hence $(\lambda z.z)\lambda y.y =_\alpha (\lambda z.z)\lambda z.z$ (as pre-terms) and so $(\lambda z.z)\lambda y.y = (\lambda z.z)\lambda z.z$ (as $\lambda$-terms).

In definition 1.1.22 (substitution) no cases are missing. Indeed, this is a definition on $\lambda$-terms (which are classes of equivalence of pre-terms, modulo $\alpha$-equivalence), hence when you write $(\lambda y.N)[x := M]$ (i.e. $([\lambda y.N]_\alpha)[x := [M]_\alpha]$), you can always suppose without loss of generality that $y$ is not a free variable in $M$ and is different from $x$. This is true because if you choose a pre-term $\lambda y.N$ where $y = x$ or $y \in FV(M)$ as a representative of the $\lambda$-term $\lambda y.N$ (i.e. of $[\lambda y.N]_\alpha$), then you can always take a pre-term $\lambda z.N[y:=z]$ where $z \notin FV(N) \cup (M) \cup \{x\}$, which is $\alpha$-equivalent to the pre-term $\lambda y.N$ and hence is another representative of the same $\lambda$-term $\lambda y.N$ (i.e. $[\lambda y.N]_\alpha$).

Concretely, apply for instance definition 1.1.22 (substitution for $\lambda$-terms) to $(\lambda y. yx)[x := yy]$. Now, $y \in FV(yy)$ so apparently we cannot apply definition 1.1.22 (which requires that the abstracted variable should not be in $FV(yy) \cup \{x\}$). However, since $\lambda y.yx$ is a $\lambda$-term, then $\lambda y. yx = \lambda z.zx$ as $\lambda$-terms (i.e. $\lambda y. yx =_\alpha \lambda z.zx$ as pre-terms) and $z \notin FV(yy) \cup \{x\}$, thus we can apply definition 1.1.22:
\begin{align} (\lambda y. yx)[x := yy] = (\lambda z. zx)[x := yy] = \lambda z.z(yy) \end{align}