I want to find the Block form matrix of the unipotent and flag subgroups $N_K$ and $G/P_K^-$ of type $B$ if we take a subset $K$ of the Dynkin diagram set $I$. For example let's take $SO_7$, which is of type $B_3$, and $K=\{1,2\}$ and, of course, $I=\{1,2,3\}$. How do I formulate these two matrices? I know how to do it if we deal with type $A$. But other than that I do not know a method that gives the explicit matrices.
Is it just (for the unipotent)
\begin{pmatrix} 1 & 0 & \ast & \ast & \ast & \ast & \ast \\ 0 & 1 & \ast & \ast & \ast & \ast & \ast \\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1\\ \end{pmatrix} I am not sure..
The first thing to notice in this case is the flags that we are seeing are not just any old flags as they are in the case of $G = SL_n$. Parabolic subgroups of $SO_n$ stabilise flags of isotopic subspaces. That is to say those which are null for the bilinear form.
Now if $P$ stabilises $V$ it also stabilises $V^\perp$. So we are looking at flags of the form $$V_1 \leq V_2 \leq \cdots \leq V_k \leq V_k^\perp \leq \cdots \leq V_2^\perp \leq V_1^\perp. $$
This gives us a basic block structure similar to the $SL_n$ case. Note that for even $n$ it is possible for the largest subspace in the flag to be its own orthogonal subspace if it is of dimension $n/2$, but this won't happen for $n$ odd since the maximum dimension will be $(n-1)/2$.
Now not all matrices of this block structure will be in in $SO_n$ but I will leave the rest up to you to work out. Note that our usual definition of orthogonal matrix no longer applies since we have changed our basis. So either we should compute what the new condition is or experiment to see what conditions we must have on the blocks.
I don't recall the exact nature of the answer but I believe we have some sort of symmetry across the off-diagonal (this may depend on choosing the basis adapted to some maximal flag containing our partial flag).
Edit: let's calculate your example. If I understand your root ordering correctly, we are looking at flag generated by an null line inside a null plane.
Thus the naive block structure will be $$\begin{pmatrix} * & * &*&*&*&*&* \\ 0& * &*&*&*&*&* \\ 0& 0 &*&*&*&*&* \\ 0& 0 &*&*&*&*&* \\ 0& 0 &*&*&*&*&* \\ 0& 0&0&0&0&*&* \\ 0 & 0 &0& 0 &0&0&* \\ \end{pmatrix}$$
Here we have chosen a basis $v_1, v_2, v_3, v_0,v_{-3}, v_{-2},v_{-1}$ which are null vectors (except for $v_0$) and moreover $(v_i,v_j) =0$ unless $j=-i$. This is exactly the kind of basis we need to diagonalise a Cartan subalgebra, by the way (for $n$ even we don't need a non-null vector $v_0$). For ease we can assume $(v_i,v_{-i}) =1$ for $i = 1,2,3$.
From here it's probably easiest to look at the Lie algebra: $$\mathfrak{so}(7) \cong \bigwedge^2\mathbb{C}^7$$ via $$ a\wedge b \mapsto (a,\cdot)b - (b,\cdot)a.$$ So for example $v_1\wedge v_2$ sends $v_{-1}$ to $v_2$, $v_{-2}$ to $-v_1$ and is zero on the rest of our basis. In particular the matrix for $v_1\wedge v_2$ looks like:
$$\begin{pmatrix} 0 & 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$
Indeed we should see that we have an anti-symmetry along the off-diagonal for any element so those off-diagonal elements will be $0$. We can also by quick calculation show that our parabolic subalgebra here is of the form $ L\wedge {C}^n + V\wedge L^\perp + \bigwedge^2V^\perp$ (note these sums are not direct and there are many different ways to write this). More importantly the nilradical is $L\wedge L^\perp + V\wedge V^\perp + \bigwedge^2V$ or as a matrix:
$$\begin{pmatrix} 0 & a & b & c & d & e & 0 \\ 0 & 0 & f & g & h & 0 & -e \\ 0 & 0 & 0 & 0 & 0 & -h & -d \\ 0 & 0 & 0 & 0 & 0 & -g & -c \\ 0 & 0 & 0 & 0 & 0 & -f & -b \\ 0 & 0 & 0 & 0 & 0 & 0 & -a \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$
We just need to take the exponential to this to find the general form of the unipotent matrices. I tried to get WolframAlpha to compute what that would look like but it wasn't a fan. The exponential series will terminate after 4 terms however so it should be possible to compute on Matlab or similar.
Note we definitely can't change our basis to write it in the form you specified since the dimension of the nilradical (and thus of the unipotent radical of the parabolic subgroup) appears to be $8$ not $10$.