Trouble Replicating Proof of The Lambda Calculus Fixed Point Theorem
$\newcommand{\Y}{\mathsf{Y}}$

Did you leave out any parentheses by any chance? We should take $\Y = (\lambda  u x . x (u u x)) (\lambda  u x . x (u u x))$. Note that $x u u x$ is the same as $(x u u) x$, which is not the same as $x (u u x)$.

We use the abbreviations $V = (\lambda x y_1 \ldots y_n . Z)$ and $X = \Y V$, as in your question. Then we calculate:
\begin{align*}
\Y V
&amp;= (\lambda  u x . x (u u x)) (\lambda  u x . x (u u x)) V \\
&amp;= (\lambda x . x ((\lambda  u x . x (u u x)) (\lambda  u x . x (u u x)) x)) V \\
&amp;= V ((\lambda  u x . x (u u x)) (\lambda  u x . x (u u x)) V) \\
&amp;= V (\Y V)
\end{align*}
and hence
\begin{align*}
X y_1 \ldots y_n
&amp;= \Y V y_1 \ldots y_n \\
&amp;= V (\Y V) y_1 \ldots y_n \\
&amp;= (\lambda x y_1 \ldots y_n . Z) (\Y V) y_1 \ldots y_n \\
&amp;= (\lambda x y_1 \ldots y_n . Z) X y_1 \ldots y_n \\
&amp;= [X/x, y_1/y_1, \ldots, y_n/y_n] Z \\
&amp;= [X/x] Z.
\end{align*}