What is the Jordan canonical form of $A^{2}$ if we know that of $A$?

Everything works blockwise, so you can simply assume that $A$ is one Jordan block...

So let $A=J_n(\lambda)$, which we can write as $\lambda I+N$ with $N=J_n(0)$. Then $A^2=\lambda^2I+2\lambda N+N^2$. The matrix $N'=2\lambda N+N^2$ is nilpotent and (because $\lambda\neq0$) has rank $n-1$, so it is conjugate to $N$. It follows that $A^2$ is conjugate to $\lambda^2I+N=J_n(\lambda^2)$.