The sentence "From here we could chose the second column of the matrix..." is ambiguous. You cannot choose $b$ and $d$, they are given. You need to prove that they are $\sin$ and $\cos$.
For example, since
$$
a^2+c^2=1,\\
d^2+b^2=1,
$$
the points $(a,c)$ and $(d,b)$ are on the unit circle and can be parameterized as
$$
(a,c)=(\cos\theta,\sin\theta),\\
(d,b)=(\cos\phi,\sin\phi).
$$
Now another two equations that you know are
$$
ab+cd=0 \quad\Leftrightarrow\quad\cos\theta\sin\phi+\sin\theta\cos\phi=\sin(\theta+\phi)=0,\\
ad-bc=1\quad\Leftrightarrow\quad\cos\theta\cos\phi-\sin\theta\sin\phi=\cos(\theta+\phi)=1.
$$
It gives $\theta+\phi=2\pi k$, which makes
$$
b=\sin(2\pi k-\theta)=-\sin\theta,\\
d=\cos(2\pi k-\theta)=\cos\theta.
$$
The matrices $O_{jk}$ are so-called Givens rotations. The goal is to show that you can transform the matrix $O$ into a upper triangular matrix with non-negative diagonal elements by multiplying (from the left) with suitable Givens rotation matrices. Those multiplications preserve orthogonality and determinant, from which we can conclude that the resulting upper triangular matrix is actually the identity. This shows that $O$ can be transformed into the identity by applying a series of Givens rotations, which in turn means that $O$ itself is the product of Givens rotation matrices (namely the inverse Givens rotations in opposite order).
The series of Givens rotations can be obtained as follows (the current values of the elements of the transformed matrix are denoted as $a_{jk}$):
For each $k,\;1\leq k <n,$ perform the following:
As long as there is a row $j>k$ with $a_{jk}\neq 0,$ find $\theta$ and $r\geq 0$ such that $r\cos\theta = a_{kk}$ and $r\sin\theta = a_{jk}.$ Multiply your matrix (from the left) with $O_{jk}$ using this $\theta.$ Afterwards, we will have $a_{kk}=r$ and $a_{jk} = 0$
This multiplication can change values only in the $j$th and $k$th column and in the $j$th and $k$th rows. All elements in the $j$th and $k$th rows on the left of the $k$th column have been made $0$ before. Therefore they keep their value $0$.
In the $k$th column, the rotation only changes $a_{kk}$ and $a_{jk}$, the latter of which is now also $0$.
After finishing this process, we have created an upper triangular matrix with non-negative diagonal elements, which is orthogonal and has determinant $1$, that is, the identity. Now "unroll" the rotations to obtain your original matrix $O$.
Best Answer
Actually, this is not the general form. It's the general form of the $2\times2$ orthogonal matrices with determinant $1$; there are also those with determinant $-1$.
Anyway, what you're after are those matrices $\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]$ such that $\left[\begin{smallmatrix}a&c\\b&d\end{smallmatrix}\right]\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]=\left[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right]$. But this means that$$\left\{\begin{array}{l}a^2+b^2=1\\c^2+d^2=1\\ab+cd=0\end{array}\right.$$The first two equations mean that $(a,b)$ and $(c,d)$ have norm $1$, whereas the third one means that $(a,b)$ and $(c,d)$ are orthogonal. Since $\bigl\|(a,b)\bigr\|=1$, $(a,b)=(\cos\theta,\sin\theta)$, for some $\theta$. And, since $(c,d)$ is orthogonal to $(a,b)$ and since it also has norm $1$, it is equal to $\pm(-\sin\theta,\cos\theta)$. Therefore, the general form is$$\begin{bmatrix}\cos\theta&\mp\sin\theta\\\sin\theta&\pm\cos\theta\end{bmatrix}.$$Replacing $\theta$ by $-\theta$, one gets the form that you mentioned.