Enumerative interpretation of generalized $q$-hockey stick identity Pascal's rule
$$ \binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k} \tag{1} $$
may be used recursively to obtain the hockey stick identity
$$ \binom{n+1}{k+1}=\binom{n}{k}+\binom{n-1}{k}+\cdots+\binom{k}{k}. \tag{2} $$
The reason for the name is that if all these binomials are highlighted in Pascal's triangle, they form what looks like a hockey stick. This is a special case of a more general identity,
$$ \binom{n+1}{r+s+1}=\sum_{a+b=n}\binom{a}{r}\binom{b}{s}, \tag{3} $$
where $\binom{a}{r}\binom{b}{s}$ counts the subsets of $\{1,\cdots,n+1\}$ where the element which separates $r$ terms on the left and $s$ terms on the right within the subset happens to separate $a$ terms on the left and $b$ terms on the right in the whole set (where we interpret left/right relative to the usual linear ordering of the numbers $1,\cdots,n+1$).
For $q$-binomials, Pascal's rule states
$$ \left[\begin{array}{c} n \\ k \end{array}\right] = \left[\begin{array}{c} n-1 \\ k-1 \end{array}\right]+q^k\left[\begin{array}{c} n-1 \\ k \end{array}\right] \tag{4} $$
which can be used recursively to obtain
$$ \left[\begin{array}{c} n+1 \\ k+1 \end{array}\right]= \left[\begin{array}{c} n \\ k \end{array}\right] + q^{k+1} \left[\begin{array}{c} n-1 \\ k \end{array}\right]+q^{2(k+1)} \left[\begin{array}{c} n-2 \\ k \end{array}\right]+\cdots \tag{5} $$
By hand, I drew out a hockey stick in $q$-Pascal's triangle, and used Pascal's rule to move one line up in the triangle. Guessing the pattern, I conjecture a generalized $q$-hockey stick identity:
$$ \left[\begin{array}{c} n+1 \\ r+s+1 \end{array}\right] = \sum_{a+b=n} q^{(b-1)(r+1)}\left[\begin{array}{c} a \\ r \end{array}\right] \left[\begin{array}{c} b \\ s \end{array}\right] \tag{6} $$
Presumably one could write the generating function proof for $(2)$ (multiply both sides by $x^n$ and sum over $n$) and then translate that into one for $(6)$, which would be interesting to see in an answer. But my major interest is:
Question. Does $(6)$ have an enumerative interpretation with objects associated to $\mathbb{F}_q^n$?
(Compare with this enumerative interpretation of the $q$-analog of the Chu-Vandermonde convolution identity.)
 A: The answer is yes.
Suppose we wish to examine an $(r+s+1)$-subspace $V\le \mathbb{F}_q^{n+1}$. Define $d_k:=\dim(V\cap\mathbb{F}_q^k)$. Note that this is monotonic ($d_1\le d_2\le \cdots$) and each $d_i$ can be at most $1$ greater than the previous $d_{i-1}$. Let $b$ be such that $d_b=s$ and $d_{b+1}=s+1$. Define $A=V\cap\mathbb{F}_q^b$ and $A'=V\cap\mathbb{F}_q^{b+1}$.
There are $\large[\begin{smallmatrix} b \\ s\end{smallmatrix}]$ choices for $A$. There is some $v'\in A'\setminus A$, which necessarily has nonzero $(b+1)$st coordinate, wlog $v'$ can be rescaled so the $(b+1)$st coordiante is $1$, and the first $b$ coordinates are uniquely determined mod $A$, i.e. it is some $v\in\mathbb{F}_q^b/A$, of which there are $q^{b-s}$ choices for.
Consider the projection onto $\mathbb{F}_q^{n+1}/\mathbb{F}_q^{b+1}$; call $B$ the image of $V$. There are $\large[\begin{smallmatrix} a \\ r \end{smallmatrix}]$ choices for $B$, where we set $a=n-b$. For any element of $B$, the corresponding vector in $V$ has its first $b+1$st coordinates uniquely determined mod $A'$, i.e. $V$ is determined by a linear map $B\to \mathbb{F}_q^{b+1}/A'$, with $q^{r(b-s)}$ possible choices.
The product of these choice counts is ${\large [\begin{smallmatrix} b \\ s\end{smallmatrix}]} q^{b-s} {\large [\begin{smallmatrix} a \\ r \end{smallmatrix}]} q^{r(b-s)}$. Summing over possible $a,b$ yields
$$ \left[\begin{array}{c} n+1 \\ r+s+1 \end{array}\right] = \sum_{a+b=n} q^{(b-s)(r+1)}\left[\begin{array}{c} a \\ r \end{array}\right] \left[\begin{array}{c} b \\ s \end{array}\right]. $$
