Let $m_3\in so(3)$ and say
$$ m_3= \begin{bmatrix} 0 & -6 & 5 \\ 6 & 0 & -4 \\ -5 & 4 & 0 \end{bmatrix} $$
It is well known that $\exp$ is many-to-one, hence if $n_3=\log(\exp(m_3))$, then $n_3\ne m_3$.
Now let $\overline m_3$ be the compact representation of $m_3$, hence
$$ \overline m_3= \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix} $$
Now intuitively $\overline m_3$ can be interpreted as a rotation vector and getting their principal value is easy. e.g. the principal value for $\overline m_3$ is:
$$ \overline p_3=\frac{\overline m_3}{\left\| \overline m_3 \right\|}(\left\|\overline m_3 \right\| \pm 2k\pi) $$
with $k\in\mathbb{N}$, such that $\left\|\overline m_3 \right\| \pm 2k\pi\in]-\pi, \pi]$.
Now it seems that for higher dimensions, the behavior is not so obvious and the $2\pi$ periodicity is only in the eigen values (my gut feeling for this intuition comes from here).
So in higher dimensions, is there a way to compute the principal value of an element?
Hence say $m_6 \in se(3)$, with:
$$ m_6= \begin{bmatrix} 0 & -6 & 5 & 1 \\ 6 & 0 & -4 & 2 \\ -5 & 4 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$
How to compute the principal value of $\overline m_6$?
Your post is very unclear.
EDIT 1. If $m\in so(3)$, then $spectrum(m)=\{0,ia,-ia\}$ where $a>0$ (if $a=0$ then $m=0$) and $m=Udiag(0,ia,-ia)U^*$ where $U$ is unitary. In general, there is $k\in\mathbb{Z}$ s.t. $a+2k\pi\in(-\pi,\pi)$.
If you choose the principal $\log$, then $\log(e^m)=Udiag(0,ia+2ki\pi,-ia-2ki\pi)U^*$, that is
$\log(e^m)=(1+\dfrac{2k\pi}{a})m$.
Remark. If $a\in \pi+2\pi\mathbb{Z}$, then $\log(e^m)$ does not exist.
Your $M=m_6=\begin{pmatrix}m&u\\0&0\end{pmatrix}$ has a double zero eigenvalue and is not diagonalizable.Then the relevant calculation is a little more complicated.
Since $M$ is similar to $diag(J_2,ia,-ia)$, it is not difficult to see that
$\log(e^M)$ is similar to $diag(J_2,ia+2ki\pi,-ia-2ki\pi)$.
Then $\log(e^M)=M-\dfrac{2k\pi}{a^3}M^3$.
Edit 2. When you calculate $\log(e^m)$, you obtain the value of $x=1+2k\pi/a$ and you deduce the principal value $p=xm$.
For the parameters of M, take $u$ and calculate $\log(e^m)$ as above.