Let $V$ be a finite dimensional vector space over a field $F$, $A$ and $B$ subspaces of $V$ such that $V=A \oplus B$. I want to show that it does not follow that $V^* = A^* \oplus B^*$.
Proof attempt: Assume that $A\neq V$. Let $f\in A^*$, then $f:A\to F$. If $g\in V^*$, then $g:V\to F$. Since $f$ and $g$ have different domains, we can't conclude that $f\in V^*$. So $A^*$ is not a subset of $V^*$ and the result follows.
On
It depends on how you interpret $\oplus$.
The dual $V^*$ is indeed isomorphic to the (external) direct sum $A^*\oplus B^*$. Indeed, if $(f,g)\in A^*\oplus B^*$, we can define $$ \overline{(f,g)}\colon V\to F \qquad \overline{(f,g)}(a,b)=f(a)+g(b) $$ and $\overline{(f,g)}$ is linear. Suppose $\overline{(f,g)}$ is the zero map. Then, for all $a\in A$, $$ \overline{(f,g)}(a,0)=f(a)=0 $$ and so $f=0$; similarly, $g=0$. Therefore the map $(f,g)\mapsto\overline{(f,g)}$ is an injective linear map $A^*\oplus B^*\to V^*$. Since domain and codomain have the same dimension, the map is also surjective.
It depends on what you mean by that equality, and how do you define the direct sum (internal or external constructions may change the result, if you want an equality of sets). What is certainly true is that $V^* \simeq A^* \oplus B^*$.
Recall that a linear map $h : A \oplus B \to W$ is uniquely determined by maps $f: A \to W$ and $g: B \to W$. That is, given two such maps, we can construct $h(a,b) := f(a)+g(b)$, and this is the only such linear function that verifies $h(a,0) = f(a)$ and $h(0,b) = g(b)$.
Now, every element of $V^* = (A \oplus B)^*$ is a linear functional $\phi : A \oplus B \to \mathbb{F}$ and so it is in bijective correspondence with maps $\phi(-,0): A \to \mathbb{F}$ and $\phi(0,-) : B \to \mathbb{F}$. That is, the function
$$ \Gamma: V^* \rightarrow A^* \oplus B^* \\ \ \phi \mapsto (\phi(-,0),\phi(0,-)) $$
is bijective. Moreover, this application is linear and so it is an isomorphism.