I just want to make sure I am understanding the definitions I'm learning correctly.
Suppose we have an isomorphism of schemes $\pi: X \to Y$, and suppose we have some affine cover of $X$, $\{U_i\}$. Is it true that $\pi|_{U_i}$ is an isomorphism for each $i$?
Yes. The restriction of a homeomorphism is a homeomorphism so that $\pi|_{U_i}:U_i\to \pi(U_i)$ is a homeomorphism. One just needs to check that it is an isomorphism on the structure sheaves, but an isomorphism of sheaves is determined at the stalk level.
Denote $\mathcal{O}_{U_i}=\mathcal{O}_X|_{U_i}$ and $\mathcal{O}_{\pi(U_i)}$ analogously for $\mathcal{O}_{Y}|_{\pi(U_i)}$. Given $x\in U_i$ and $\pi(x)\in \pi(U_i)$, then the stalk map $\mathcal{O}_{\pi(U_i),y}\to \mathcal{O}_{U_i,x}$ of $\pi|_{U_i}$ agrees with the stalk map $\mathcal{O}_{\pi(U_i),y}\to \mathcal{O}_{U,x}$ for $\pi$. So, the map $\pi|_{U_i}$ is also an isomorphism on stalks. Hence, we are done.