Lambda calculus :Will there be any x left after first substitution?
After $[Q/x]M$ we no longer find free occurrences of $x$ ... unless $x$ is free in $Q$! The simpler example is $Q=x$, where $[x/x]M=M$ can have free occurrences of $x$.

In such case, the free $x$s due to $Q$ can be replaced by $[P/x]$ in your example.

E.g. $[z/x][(\lambda y.\ x)/x](x(yx)) = ((\lambda y.\ z)(y(\lambda y.\ z)))$.

According to (e) we have indeed

$$
\begin{array}{l}
[z/x][(\lambda y.\ x)/x](x(yx)) \\
= [[z/x](\lambda y.\ x)/x](x(yx))\\
= [(\lambda y.\ z)/x](x(yx))\\
= ((\lambda y.\ z)(y(\lambda y.\ z)))
\end{array}
$$

Finally, note that when $x$ is not free in $Q$, we have
$[[P/x]Q/x]M = [Q/x]M$ as an immediate consequence of $[P/x]Q=Q$.