This is Lemma 1.3.14 from Scharlau's book Quadratic and Hermitian Forms.
3.14 Lemma. Let $\varphi$, $\psi$, $\varphi_1$, $\psi_1$ be bilinear spaces.
(i) $\psi\perp\varphi \cong \varphi\perp\psi$.
(ii) If $\varphi\cong\varphi_1$, $\psi\cong\psi_1$, then $\varphi\perp\psi \cong \varphi_1\perp\psi_1$.
(iii) $\varphi\perp\psi$ is regular if and only if $\varphi$ and $\psi$ are regular.
(iv) If $B$ is the matrix of $\psi$ and $B'$ is the matrix of $\psi$, then the orthogonal sum $\varphi\perp\psi$ has the matrix $$\begin{pmatrix} B & 0 \\ 0 & B' \end{pmatrix}.$$
In the proof the proof is omitted with the comment that it's easy.
Definition of external orthogonal sum is given in Definition 1.3.13. The underlying vector space is $V\oplus V'$ and the bilinear for $b\perp b'$ is defined by $$(b\perp b')((x,x'),(y,y'))=b(x,y)+b'(x',y').$$
First two result are clear since bilinear form maps to commutative field $K$ but for regularity if both form are regular then external sum is regular is easy to show but what about other way and about (iv).
This is about the opposite direction for (iii).
$(V,b)$ is regular iff $V^\bot=\{0\}$; where $V^\bot= \{x\in V; (\forall v\in V) b(x,v)=0\}$.
We want to show that if $(V\oplus W,b\perp b')$ is regular, then $(V,b)$ is regular.
Suppose that $x\in V^\bot$. This means that $b(x,v)=0$ for every $v\in V$.
Then we also have $b\perp b'((x,0),(v,w))=0$ for every $v\in V$, $w\in W$ simply from $$b\perp b'((x,0),(v,w))=b(x,v).$$ This implies that $(x,0)\in (V\oplus W)^\bot$ and thus $$(x,0)=(0,0)$$ (using regularity) and therefore $x=0$.
We have shown that $V^\bot=0$, which means that $(V,b)$ is regular.
The fact that $(W,b')$ is regular can be shown similarly.
For (iv) it is enough to look at $b\perp b'$ with respect to the natural basis on $V\oplus W$ determined by the bases of $V$ and $W$. I.e., if $(e_1,\dots,e_n)$ is basis of $V$ and $(f_1,\dots,f_k)$ is basis of $W$, you simply take basis consisting of elements of the form $(e_i,0)$ and $(0, f_j)$, where $i=1,\dots,n$, $j=1,\dots,k$.