This is a nice question. I have never seen this before.

Let us write
$$\mathcal B_u = \{ V_0 \subset V_1 \subset \cdots \subset V_{n-1} \subset V_n = \mathbb C^n : u V_i \subset V_i \}. $$
Let $ \lambda $ be the Jordan type of $ u $. Then we can partition $ B_u $ into locally closed subsets depending on the Jordan type of $ u \rvert_{V_{n-1}} $; such a Jordan type $ \mu$ must necessarily be obtained by removing one box from $ \lambda$. So we have
$$ \mathcal B_u = \bigsqcup_\mu B_u^\mu \quad B_u^\mu = \{ V_\bullet : u\rvert_{V_{n-1}} \text{ has Jordan type $ \mu$ } \}.$$
Let $ G(n-1, n)_u $ denote the set of all $n-1$-dimensional $u$-invariant subspaces of $ \mathbb C^n $ and partition $ G(n-1,n)_u $ into locally closed subsets $ G(n-1,n)_u^\mu $ according to the Jordan type of the restriction of $ u $ to the subspace.

We have $ \mathcal B_u \rightarrow G(n-1,n)_u $ and $ \mathcal B^\mu_u $ is the preimage of $ G(n-1,n)^\mu_u$. Note that the fibre of $ \mathcal B_u \rightarrow G(n-1,n)_u $ is $ \mathcal B_{u\rvert_{ V_{n-1}}}$.

Now, here are two facts that I don't see right away, but which must be
true:

- Each piece $ \mathcal B_u^\mu $ has the same dimension.
- $ G(n-1,n)^\mu_u $ is irreducible and simply-connected (or at least that the bundle $\mathcal B^\mu_u \rightarrow G(n-1,n)^\mu_u$ is topologically trivial on components).

**Edit:** I think these facts should be contained in the classic paper by Spaltenstein https://www.sciencedirect.com/science/article/pii/S138572587680008X

Assuming these facts, we get
$$H(\mathcal B_u) = \bigoplus_\mu H(\mathcal B_u^\mu) = \bigoplus_\mu H(\mathcal B_{u(\mu)})$$
where again the direct sum ranges over all partitions made by deleting one box from $\lambda$ and where $ u(\mu)$ denotes a unipotent element of Jordan type $ \mu$, and $ H(X) $ denotes top (co)homology. This gives the desired decomposition.