How do I have to interpret the input to the OCC flag in Molpro?
You are correct in your interpretation of the OCC flag (that command corresponds to $3$ electrons in $\mathrm{a_g}$ orbitals, $1$ in $\mathrm{b_{3u}}$, $1$ in $\mathrm{b_{2u}}$, and $2$ in $\mathrm{b_{1u}}$). (Because these are orbitals, not overall wavefuction symmetries we use lowercase letters)

Your confusion seems to be arising from visualizing the orbitals which you are including. Remember that these representations refer to the symmetry of the molecular orbital, not atomic orbitals. $\mathrm{a_g}$ refers to the set of orbitals which have no phase change under all of the symmetry operations, so that corresponds to the $\sigma(\ce{1s})$, $\sigma(\ce{2s})$, and $\sigma(\ce{p_z})$, orbitals. Similarly, $\mathrm{b_{1u}}$ refers to the filled $\sigma^*(\ce{1s})$, $\sigma^*(\ce{2s})$. Finally the $\mathrm{b_{3u}}$, and $\mathrm{b_{2u}}$ orbitals refer to the filled $\pi(\ce{p_x})$ and $\pi(\ce{p_y})$ orbitals. 

As we know, the occupation of $\ce{N2}$ is: 
$$
  \ce{\sigma(1s)^2 \sigma(2s)^2 \sigma^*(1s)^2 \sigma^*(2s)^2 
  \pi(p_x)^2 \pi(p_y)^2 \sigma(p_z)^2},
$$ 
and using the $D_\mathrm{2h}$ point group we can rewrite this as: 
$$\mathrm{1a_g^2\ 2a_g^2\ 1b_{1u}^2 \ 2b_{1u}^2 \ 1b_{3u}^2\ 2b_{3u}^2\ 3a_g^2}.$$
Hopefully, this rewriting of the occupation pattern in the chosen symmetry makes the choice of orbitals clear.

To really understand which orbitals fall under which irreducible representation you need to visualize them (or draw them in the case of a dimer) and then pull out the character table for the point group. 

You can find a $D_\mathrm{2h}$ character table on the homepage of Dr. Achim Gelessus (Jacobs University Bremen).