When is $V=U\oplus U^{\perp}$?

Construction

Here's a recipe to construct "bad" incomplete spaces:

1. Start with a Hilbert space $\dim\mathcal{H}=\infty$.
2. Choose a normalized vector $e_0$.
3. Extend it to an ONB $\mathcal{E}\owns e_0$.
4. Fix the independent vector $b_0:=e_0+\sum_{k=1}^\infty\frac{1}{k}e_k$.
5. Extend this to a Hamel basis $\mathcal{B}\supseteq\mathcal{E}$ with $\mathcal{B}\owns e_0,b_0$.
6. Rip it off to get an orthonormal system $\mathcal{S}:=\mathcal{E}\setminus\{e_0\}$.
7. Rip it off to get a linear independent system $\mathcal{L}:=\mathcal{B}\setminus\{e_0\}$.
8. Span your incomplete space $X:=\langle\mathcal{L}\rangle$.

Then the orthonormal system is maximal $\mathcal{S}^\perp=(0)$ but not an ONB $\overline{\langle\mathcal{S}\rangle}\neq X$.

Example

For your query then you can split $\mathcal{S}=\mathcal{S}_1\sqcup\mathcal{S}_2$ to get subspaces $U_1:=\overline{\langle\mathcal{S}_1\rangle}$ and $U_2:=\overline{\langle\mathcal{S}_1\rangle}$. These are orthogonal complements to each other but don't reduce the space $X\neq U_1\oplus U_2$.

Moreover you see that any combination of dimension and codimension can appear to be bad in an incomplete space.

I'm adding another answer, now that you've changed the question.

Assume that $W$ is finite-dimensional with $W^{\perp} \subseteq U$. The goal is to show that $U\oplus U^{\perp} = V$. All '$\oplus$' decompositions are orthogonal in what follows.

Because $W$ is finite-dimensional then $V=W\oplus W^{\perp}$ which can be seen by choosing any basis of $W$ and using Gram-Schmidt to find an orthonormal basis $\{ e_{1},\cdots,e_{n}\}$ of $W$. Then every $v\in V$ can be written as $$ v = \left(v-\sum_{j=1}^{n}(v,e_{j})e_{j}\right)+\sum_{j=1}^{n}(v,e_{j})e_{j}. $$ So $W^{\perp}\oplus W=V$. Assuming that $W^{\perp}\subseteq U$, it follows that every $u \in U$ can be written as $u=w_{\perp}+w$ where $w_{\perp}\in W^{\perp}$ and $w\in W$. Because $w_{\perp} \in W^{\perp}\subseteq U$, then $u-w_{\perp}=w \in U$, which gives the decomposition $$ U = W^{\perp}\oplus(U\cap W). $$ Because $U\cap W$ is a finite-dimensional subspace of $W$, then it follows that there exists a finite-dimensional subspace $U'$ such that $(U\cap W)\oplus U'=W$; $U'$ is found by completing an orthonormal basis of $U\cap W$ to one for $W$. Therefore $(U\cap W)\oplus U'=W$. Finally, $$ \begin{align} U\oplus U' & = (W^{\perp}\oplus (U\cap W))\oplus U' \\ & = W^{\perp}\oplus((U\cap W)\oplus U') \\ & = W^{\perp}\oplus W = V. \end{align} $$ This is enough to give $U'=U^{\perp}$ and $U\oplus U^{\perp}=V$.

The hypothesis that $U$ has finite co-dimension is not enough. The hypothesis that $U$ is closed is not enough. I'll give you an example where both hold and it's still not true that $V=U\oplus U^{\perp}$.

Let $V$ the be the linear space generated by finite linear combinations of standard basis elements $\{ e_{j}\}_{j=1}^{\infty}$ of $l^{2}$. Define a linear functional $\Phi$ on $V$ by $$\Phi(x)=\sum_{j=1}^{\infty}\frac{1}{j}(x,e_{j}).$$ $\Phi$ is a bounded linear function on $V$ and, therefore, $U=\mathcal{N}(\Phi)$ is closed in $V$; and $U$ is of co-dimension $1$ in $V$ because $\Phi$ is not the $0$ functional on $V$. Therefore $V \ne U$. However, $U^{\perp}=\{ x \in V : (x,u)=0 \mbox{ for all } u \in U\}=\{0\}$ which means that $V \ne U = U\oplus U^{\perp}$.