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Until otherwise mentioned, $V$ is a finite dimensional vector space over a field $F$ and $f$ is a Bilinear form on $V$.
We will use $L^2(V, F)$ to denote the set of all the bilinear forms on $V$ and $\ann(W)$ will denote the annihilator of a subspace $W$ of $V$.
Fundamentals
Definition.
For each $f\in \mc L^2(V;\ F)$ define $L_f:V\to V^*$ as
$$
L_f(u)=g_u,\quad \forall u\in V
$$
where $g_u\in V^*$ is define as
$$
g_u(v)=f(u,v),\quad\forall v\in V
$$
Similarly, for each $f\in \mc L^2(V,F)$ define $R_f:V\to V^*$
as
$$
R_f(v)=h_v,\quad \forall v\in V
$$
where $h_v\in V^*$ is defined as
$$
h_v(u)=f(u,v),\quad\forall u\in V
$$
It can be seen that $L_f$ and $R_f$ are linear transformations for each $f\in \mc L^2(V;\ F)$.
The following theorem is immediate:
Theorem.
Let $f\in \mc L^2(V;\ F)$.
Then $\rank L_f=\rank R_f$.
Proof.
Identifying $V^{**}$ with $V$, it can be easily seen that $L_f^t=R_f$.
$\blacksquare$
Definition.
Let $f\in \mc L^2(V;\ F)$.
Then the rank of $f$ is defined as
$$\rank f=\rank L_f=\rank R_f$$
We say say that $f$ is non-degenerate if $\rank f=\dim V$.
More generally, we say that $f$ is non-degenerate on a subspace $W$ of $V$ if $\rank L_{f|W}=\dim W$.
Orthogonality
Let $f\in \mc L^2(V;\ F)$.
Let $u$ and $v$ be two vectors in $V$.
We say that $u$ is perpendicular to $v$, written $u\perp v$, if $f(u,v)=0$.
Given subspace $W$ of $V$, we define the orthogonal complement of $W$ as $W^\perp=\set{u\in V:\ w\perp u\text{ for all } w\in W}$.
Note that $v\in W^\perp$ if and only if $R_fv\in \ann(W)$.
Theorem.
Let $f$ be a non-degenerate bilinear form on a finite dimensional vector space $V$.
Then for any subspace $W$ of $V$, we have $\dim(W)+\dim(W^\perp)=\dim(V)$.
Proof.
Since $f$ is non-degenerate, we have $R_f:V\to V^*$ is an isomorphism.
As just commented, we have $v\in W^\perp$ if and only if $R_fv\in W^\perp$.
Thus $W^\perp=R_f^{-1}(\ann(W))$.
Since $R_f$ is an isomorphism, we have $\dim(R_f^{-1}(\ann(W)))=\dim(\ann(W))=\dim(V)-\dim(W)$.
Thus $\dim(W^\perp)=\dim(V)-\dim(W)$, and we have the desired equality.
$\blacksquare$
The conclusion is no longer true if we drop the non-degenracy condition. This is because if we construct a bilinear form for which the image of $R_f:V\to V^*$ intersects $\ann(W)$ only in $0$, then we would have $W^\perp=0$
Since $V=U\oplus U^\perp$, every vector in $V$ can be written as a sum of a vector in $U$ and a vector in $U^\perp$.
This holds, in particular, for $u\in(U^\perp)^\perp$. So, according to Axler, write
$$
u=v+w,\qquad v\in U, w\in U^\perp
$$
Now apply the assumption that $u\in(U^\perp)^\perp$ and the fact that $v\in U\subseteq(U^\perp)^\perp$ to conclude that
$$
w=u-v\in(U^\perp)^\perp
$$
as well. On the other hand, $w\in U^\perp$ by hypothesis. Therefore
$$
w=u-v\in U^\perp\cap(U^\perp)^\perp
$$
A general result about orthogonal complements is that, for every subspace $X$, $X\cap X^\perp=\{0\}$. This holds in particular for $X=U^\perp$.
Hence $w=u-v=0$, so $u=v\in U$.
Best Answer
If $U$ is a line NOT going through the origin in ${\mathbb R}^3$, then $U^{\bot}$ is a line going through the origin and perpendicular to the plane containing $U$ and the origin. That would be an example when $U \cap U^{\bot} = \emptyset$. The reason is that $U^{\bot} = (sp(U))^{\bot}$, where $sp(U)$ is the span of $U$ - the smallest linear subspace containing the vectors in $U$: if a vector $v$ is perpendicular to all vectors in $U$, then it is perpendicular to all linear combinations of vectors in $U$.