Let $U$ be a finite-dimensional vector space and $W=U\oplus U'$. Halmos asserts(in his FDVSs book) that the correspondence $(x,y) \mapsto (y,x)$ is an isomorphism between $W$ and $W'$.
I'm not sure I understood his notation correctly. Since in the finite dimensional case he does use $x$ in place of the element of $U''$ which is related to it by reflexivity I'm thinking he means that $(y,x)$ is $g \in W' ~|~ g(X,Y) = y(X)+x(Y)$. This indeed makes it an isomorphism. Is there another possible interpretation for this?
Recall that every finite dimensional $k$-vector space $X$ is canonically isomorphic to its double dual $X''$.
Moreover, if you have a direct sum of two vector spaces $V=A\oplus B$, you can consider $\{a_i\}_i$ a basis of $A$, $\{b_j\}_j$ a basis of $B$, so that $\{a_i\}_i \cup \{b_j\}_j$ is a basis of $V$. Since linear maps can be defined simply on bases, you have the canonical isomorphism $$\operatorname{Hom}(A \oplus B,k) \cong \operatorname{Hom}(A,k) \oplus \operatorname{Hom}(B,k)$$ acting as $f \mapsto (f|_A, f|_B)$, with inverse $(g,h) \mapsto ((a,b) \mapsto g(a)+h(b))$.
So $$W' = \operatorname{Hom}(U \oplus U',k) \cong \operatorname{Hom}(U,k) \oplus \operatorname{Hom}(U',k) = U' \oplus U'' \cong U'\oplus U$$ this means that $W'$ is canonically isomorphic to $W$ simply by swapping the two coordinates (i.e. precisely $(x,y) \mapsto (y,x)$).