Let's take the matrix $R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
Consider its transpose $R^\dagger = \begin{pmatrix} a^* & c^* \\ b^* & d^* \end{pmatrix}$.
Then $RR^\dagger =1$ implies that $|a|^2 + |c|^2 = 1, |b|^2 + |d|^2 = 1, ab^* + cd^* = 0,$ and $a^*b + c^*d = 0$.
The last two equations are equivalent to each other, so we have three complex equations.
If $a=a_0+ia_1, b=b_0+ib_1, c=c_0+ic_1,$ and $d=d_0+id_1$, then the above three complex equations lead to $a_0^2 + a_1^2+c_0^2+c_1^2=1, b_0^2 + b_1^2+d_0^2+d_1^2=1, a_{0}b_{0} + a_{1}b_{1} + c_{0}d_{0} + c_{1}d_{1} = 0,$ and $a_{0}b_{1} - a_{1}b_{0} + c_{1}d_{0} - c_{0}d_{1} = 0$.
Therefore, the three complex equations lead to four real equations (the first two equations have only real components, but the final equation has both real and complex components).
Therefore, the four real equations ensure that the number of real variables that parametrise the matrix $R$ drops from 8 to 4.
Now, let's consider the determinant equation $ad-bc=1$, which decomposes into $(a_{0}d_{0} - a_{1}d_{1} - b_{0}c_{0} + b_{1}c_{1}) + i(a_{1}d_{0} + a_{0}d_{1} - b_{1}c_{0} - b_{0}c_{1}) = 1,$ which decomposes into $a_{0}d_{0} - a_{1}d_{1} - b_{0}c_{0} + b_{1}c_{1} = 1,$ and $a_{1}d_{0} + a_{0}d_{1} - b_{1}c_{0} - b_{0}c_{1} = 0$.
Therefore, we have two real equations, so the number of parameters that characterise the matrix $R$ should now drop from 4 to 2. But several sources mention that the number of parameters drop from 4 to 3.
Can someone explain?
The explanation is that the determinant is already constrained to be on the unit circle. We have $|\det(R)|^2 = \det(R)\overline{\det(R)} = \det(R)\det(R^\dagger) = \det(RR^\dagger) =\det(I) = 1$.
This means that the equation $a_0d_0 - a_1d_1 - b_0c_0 + b_1c_1 = 1$ implies $a_0d_1 + a_1d_0 - b_0c_1 - b_1c_0 = 0$, and the latter equation implies $a_0d_0 - a_1d_1 - b_0c_0 + b_1c_1 = \pm 1$. In other words, the set of five real equations has a nontrivial relation, which increases the dimension of the solution set.
Another way of looking at it is that the determinant gives a homomorphism $U(2) \to S^1$, and the first isomorphism theorem then guarantees that the dimension of the kernel $SU(2)$ is the dimension of the source minus the dimension of the image.