Normal order sequencing vs applicative order sequencing
You're misreading the combinator. In these lecture notes, lambda abstraction binds tighter than application (the opposite convention is more common — it's apparent from the spacing but the author should state that explicitly). I'll use extra (redundant) parentheses in my answers to avoid any potential for ambiguity. The combinator is
$$ \mathit{Seq} = \lambda x. \lambda y. ((\lambda z.y) \: x)$$
So
$$ \begin{align}
  &amp; ((\mathit{Seq} \; (\text{display "hello"})) \; (\text{display "world"})) \\
  &amp;\qquad = (((\lambda x. \lambda y. ((\lambda z.y) \; x)) \; (\text{display "hello"})) \; (\text{display "world"})) \\
  &amp;\qquad \to^2 ((\lambda z. (\text{display "world"})) \; (\text{display "hello"})) \\
  &amp;\qquad \xrightarrow{\text{display "hello"}} (\text{display "world"}) \\
  &amp;\qquad \xrightarrow{\text{display "world"}} \mathord{?} \\
\end{align} $$
where $M \xrightarrow{\text{action}} N$ means that $M$ evaluates to $N$ performing the specified action.