This question is about (a special case of) the varieties discussed here: Does the preimage of the Slodowy slice in $T^*G/P$ have a name?.

Let $G = SL_n(\mathbb{C}), \mathfrak{g} = \mathfrak{sl}_n(\mathbb{C})$; $\mathcal{N}$ its nilpotent cone and let $z \in \mathfrak{g}$. The Springer resolution is $\widetilde{\mathcal{N}}=T^*G/B = \{ (0 \subset V_1 \subset \cdots \subset V_n), x | x V_{i} \subset V_{i-1} \}$; the natural map $\pi: \widetilde{\mathcal{N}} \rightarrow \mathcal{N}$ is a resolution of singularities.

Let $e \in \mathcal{N}$ be a nilpotent (I'm in the case where $z$ is a $2$-block nilpotent, but I'm not sure if this simplifies matters much). Using Jacobson-Morozov, pick an $\mathfrak{sl}_2$-triple $\{ e, h, f \}$; then the Slodowy slice is defined to be $S = e + Z_{\mathfrak{g}}(f)$ (where $Z_{\mathfrak{g}}(f)$ is the centralizer of $f$ in $\mathfrak{g}$).

Let $S' = S \cap \mathcal{N}$, and $\widetilde{S'} = \pi^{-1}(S') = \{ (V_{\bullet}), x | x \in S, x V_{i} \subset V_{i-1} \}$. It is known that $\widetilde{S'}$ is smooth.

My question: given a point $t = \{ (V_{\bullet}), x\} \in \widetilde{S'}$, what is $T_{t}(\widetilde{S'})$ (the tangent space to $\widetilde{S'}$ at $t$)? Since $\widetilde{S'}$ is a subvariety of $\widetilde{\mathcal{N}}$, the answer should be as a subspace of $T_t(\widetilde{\mathcal{N}}) = \mathfrak{n}_t^* \oplus \mathfrak{n}_t$, where $\mathfrak{n}_t$ is the nilpotent radical of the Borel $\mathfrak{b}_t$ corresponding to $(V_{\bullet}) \in G/B$. Here the tangent space $T_t(\widetilde{\mathcal{N}}) = \mathfrak{n}_t^* \oplus \mathfrak{n}_t$, because $\widetilde{\mathcal{N}} \rightarrow G/B$ is a vector bundle; the tangent space $T_{(V_{\bullet})}(G/B) = \mathfrak{g}/\mathfrak{b}_t \simeq \mathfrak{n}_t^*$ and the fiber above $(V_{\bullet})$ is $\mathfrak{n}_t$.