Orthogonal Complement of Direct Sum

First we should note that the direct sum $A\oplus B$ is defined only for subspaces $A$ and $B$ (and not for just any two subsets!). If $A$ and $B$ are just subsets of $H$, we can define the sum $$A+B=\{a+b:a\in A, b\in B\},$$and now if $A$ and $B$ are subspaces and if $A\cap B=\{0\}$ then the sum is said to be the "direct sum" of $A$ and $B$: $$A\oplus B=A+B.$$

And for $A\subset H$ we define $$A^\perp=\{x\in H: <x,a>=0\,\forall a\in A\},$$and note that $A^\perp=S^\perp$ if $S$ is the span of $A$ (or the closed span). At least if $A$ and $B$ both contain the origin, determining $(A+B)^\perp$ is trivial:

Trivial Exercise. If $A,B\subset H$ and $0\in A\cap B$ then $(A+B)^\perp=A^\perp\cap B^\perp$.

(Hence $(A\oplus B)^\perp=A^\perp\cap B^\perp$ in the only case in which $A\oplus B$ is defined, namely $A$ and $B$ are subspaces with trivial intersection.)

We give the proof, since the OP has expressed some skepticism: First, $0\in B$ shows that $A\subset A+B$, hence $(A+B)^\perp\subset A^\perp$. Similarly $(A+B)^\perp\subset B^\perp$, so $(A+B)^\perp\subset A^\perp\cap B^\perp$.

The other inclusion is even easier. Suppose $x\in A^\perp\cap B^\perp$. Then $$<x,a+b>=<x,a>+<x,b>=0+0=0\quad(a\in A, b\in B),$$so $x\in(A+B)^\perp$.