Any matrix $A \in SO(4)$ can be written as $$A = \exp{\left( \sum_{n=1}^6 \lambda_n B_n \right)} \qquad \left( \textrm{with }\lambda_n \in \mathbb R , \;n = 1, \cdots, 6\right)$$ where $B_n$ are the six standard basis elements for the space of $4\times 4$ anti-symmetric matrices (which are the generators of the Lie algebra $\mathfrak{so}(4)$). More explicitly, fixing a particular ordering: $$ \sum_{n=1}^6 \lambda_n B_n = \left( \begin{array}{cccc} 0 & \lambda_1 & \lambda_2 & \lambda_3 \\ -\lambda_1 & 0 & \lambda_4 & \lambda_5 \\ -\lambda_2 & -\lambda_4 & 0 & \lambda_6 \\ -\lambda_3 & -\lambda_5 & - \lambda_6 & 0 \end{array} \right) $$
I am now wondering, given a particular $A$, how can I obtain the above coefficients (in a way which is also practically feasible)?
An element $A$ of $SO(4)$ has a normal normal form: there is an orthogonal matrix $P$ such that $PAP^t$ is a block diagonal matrix with $2\times 2$ blocks, and which along the diagonal has blocks of the form $\left( \begin{array}{cc} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \\ \end{array} \right)$. Find it. If you replace replace $\alpha$ by $\alpha t$, you get a $1$-parameter group in $SO(4)$ going through your matrix: its derivative at $0$ is an element of the Lie algebra and $A$ is the exponential of a real multiple of that derivative.