Evaluation of $\beta$-Reduction with Parentheses in $\lambda$-Calculus
You are missing one important pair of parentheses, around an expression for $2$. It should be
$$\bigl(\lambda s. \lambda z. (s\ (s\ z))\bigr)\ s\ (s\ z)$$
Then we reduce by replacing $s$ with $s$ and $z$ with $(s\ z)$, obtaining
$$(s\ (s\ (s\ z)))$$



On the other hand, $\lambda s. \lambda z. (s\ (s\ z))\ s\ (s\ z)$ doesn’t reduce any further because it has no redex. It is just a $\lambda$-abstraction with variables $s,z$ and body $(s\ (s\ z))\ s\ (s\ z)$. Abstraction always goes as far right as possible.



Generally, when we have a name for expression (e.g. $2$, $myexp$,...) within some larger expression, we must surround it with parentheses when we expand it to its true lambda form, to avoid confusion regarding applications.
Just like you did with $1$ in the third row.  

You can explore $\lambda$-calculus reductions, test your ideas and see how various expressions are implemented with this interpreter.