Intersection of open affines can be covered by open sets distinguished in *both*affines

This is true. For a scheme $X$, let $f$ be a global section of $O_X$. You can construct an open set $X_f$ defined as points where $f$ doesn't vanish, i.e. $\{x \in X: f_x \notin m_x\}$, where $m_x$ is the maximal ideal of the local ring. For affine schemes, this agrees with the usual distinguished open affine $D(f)$. One can also check that if $U \subset X$ is open, then $X_f \cap U = U_{f|_U}$.

In your situation, let $x \in U \cap V$. Take a section $f$ over $U$ such that $x \in D_U(f) \subset U \cap V$ is distinguished in $U$. Take a section $g$ over $V$ such that $x \in D_V(g) \subset D_U(f)$ is distinguished in $V$. The claim is that $D_V(g)$ is also distinguished in $D(f)$, hence in $U$. This is because $D_V(g) = D_{D_U(f)} (g|_{D_U(f)})$.

You can find it as Proposition 3.2 here (Ravi Vakil's notes on algebraic geometry, very recommended!).