Consider in $(\mathbb{R}, \mathbb{R}^n, +)$ a subspace $U$. Now $W$ is a subspace of $U$. Prove that: $$U^{\perp_{\mathbb{R}^n}} + W = (W^{\perp_U})^{\perp_{\mathbb{R}^n}}$$ where the orthogonal complement of the subspace has to be taken in the explicitly mentioned vector space, towards the default inproduct of $\mathbb{R}^n$.
Does it suffice to show that
$$(U^{\perp_{\mathbb{R}^n}} + W) \oplus W^{\perp_U} = \mathbb{R}^n \land (U^{\perp_{\mathbb{R}^n}} + W) \perp W^{\perp_U}$$
$$\iff (U^{\perp_{\mathbb{R}^n}} + W) \cap W^{\perp_U} = 0 \land (U^{\perp_{\mathbb{R}^n}} + W) + W^{\perp_U} = \mathbb{R}^n \land (U^{\perp_{\mathbb{R}^n}} + W) \perp W^{\perp_U}$$
$$\iff \dim(U^{\perp_{\mathbb{R}^n}} + W) + \dim(W^{\perp_U}) = n \land (U^{\perp_{\mathbb{R}^n}} + W) \perp W^{\perp_U}$$
I have my doubts about the last equivalence: is it true that if the dimensions are equal to $n$, that the subspaces themselves are equal to $\mathbb{R}^n$ in this case? And is it true that if $(U^{\perp_{\mathbb{R}^n}} + W) \perp W^{\perp_U}$ that then automatically $(U^{\perp_{\mathbb{R}^n}} + W) \cap W^{\perp_U} = 0$?
And also, is it overkill to show that the dimensions are equal, since the sum of vector spaces is associative and thus it is very easy to see that $U^{\perp_{\mathbb{R}^n}} + (W + W^{\perp_U}) = \mathbb{R}^n$ in my second equivalence?
No, although $A\oplus B=C$ (with in your case $A=U^{\perp_{\mathbb{R}^n}} + W$, $B=W^{\perp_U}$ and $C=\mathbb{R}^n$) implies $\dim A+\dim B=\dim C$, the latter (checking the dimensions) does not imply the former. But it is true that $A\perp B$ implies $A\cap B=\{0\}$ (since any vector in the intersection has to be perpendicular to itself, forcing it to be $0$), so indeed the final condition of the three in your middle display implies the first one. And $A\cap B=\{0\}$ does imply that $\dim(A+B)=\dim A+\dim B$, so if you also know that $A+B\subseteq C$ and $\dim A+\dim B=\dim C$ (as is apparently the case here), then it does follow that $A\oplus B=C$.
And you are right that alternatively the conditions $A\perp B$ and $A+B=C$ do imply that $A\oplus B=C$ (I more or less gave the argument), so you might go along that path and not consider dimensions at all.