We can show that the sequents: $(p \to q) \lor (q \to r)$ is a tautology.
$((p \to q) \lor (q \to r))$
$\iff ((\lnot p \lor q) \lor (\lnot q \lor r))$
$\iff (\lnot p \lor \color{blue}{\bf (q \lor \lnot q)} \lor r)$
$\iff (\lnot p \lor \color{blue}{\bf\text{TRUE}} \lor r)\qquad$ by $\color{blue}{\bf LEM}$ (Either $q$ is true, or else $\lnot q$ is true.)
$\equiv \text{TRUE}$
$\therefore$ The sequent is valid as a tautology.
For a more detailed discussion of the Law of Excluded Middle (LEM), see this Wiki article
Note, one can "build up" (derive) the desired expression from only an application of the law of the excluded middle:
Start with $q \lor \lnot q$.
Introduce the disjunct $\lnot p$, to get $\lnot p \lor (q \lor \lnot q)$.
Through associativity of disjunction, we can express that as $(\lnot p \lor q) \lor \lnot q$.
Introduce another disjunct $r$, to get $(\lnot p \lor q) \lor \lnot q \lor r$, and
Through associativity of disjunction, again, we can arrive at $(\lnot p \lor q) \lor (\lnot q \lor r)$.
Now we need only apply equivalences discussed above to arrive at the proposition: $(p \to q) \lor (q \to r)$