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Use a Gelfand–Tsetlin basis in Young seminormal form.

For a standard Young tableau $T$, put $d=c_T(i+1)-c_T(i)$, using the Content of a Young-diagram cell. Choose the row-reading tableau $T_0$, and let $\ell (T)$ be the Coxeter length of the unique permutation sending $T_0$ to $T$. A Gelfand–Tsetlin basis can be chosen so that, when $R=s_{i}T$ is standard and $\ell (R)=\ell (T)+1$,If $R$ is not standard, the action is $+v_T$ for two consecutive entries in one row, and $-v_T$ for two in one column. This is one usual normalization of the Young seminormal form.

Here is a construction and proof of the normalization. Fix $v_{T_0}\ne 0$, let $P_T$ be the projection onto the tableau line, and defineThe permutation ${\pi }_T$ has a reduced expression consisting entirely of admissible swaps, by the reduced adjacent-swap path between linear extensions. At each swap the off-diagonal coefficient is nonzero. In its expansion, the only term that can reach a tableau at distance $\ell (T)$ uses all $\ell (T)$ swaps; omitting a swap gives a shorter path. Thus $v_T\ne 0$. This also makes its definition independent of a chosen reduced expression, because ${\pi }_T$ itself is fixed.

The relation $s_i{X}_i{s}_i+s_i=X_{i+1}$ forces the coefficient of $v_T$ in $s_i{v}_T$ to be $1/d$, and forces every other component to lie on the swapped line. If length increases, project the identity ${\pi }_R=s_i{\pi }_T$ onto the line for $R$. The shorter terms of ${\pi }_T{v}_{T_0}$ cannot reach $R$ in one step, so the coefficient of $v_R$ is exactly one. Applying $s_i^2=1$ then gives the reverse coefficient $1-d^{-2}$ and diagonal coefficient $-1/d$. In the nonstandard cases the same relation gives $d=\pm 1$ and the asserted scalar action. This proves the theorem rather than merely specifying pairwise scalings.

The inclusion here is inclusion of Young diagrams, not the dominance order on partitions. The skew Young diagram $\lambda /\mu$ is the set difference of their cells. Two cells are adjacent when they share an edge. It is connected when any two cells can be joined by such steps, and is a rim hook when it is connected and contains no $2\times 2$ square. Under the usual edge-adjacency convention a totally disconnected skew Young diagram has only singleton components, equivalently no two cells share an edge. A horizontal strip instead means at most one cell per column; this distinction matters for the last part of this question.

A standard skew Young tableau is a linear extension of a partially ordered set: the cells are ordered by the row and column inequalities, and the tableau lists them in increasing label order. Write the current list as $x_1,\dots ,x_k$ and let $r_j$ be the desired label of $x_j$ in $R$. Then $(r_1,\dots ,r_k)$ is the one-line notation of the unique permutation $\pi$ with $\pi T=R$.

Whenever this list of desired labels is not increasing, there is an adjacent descent $r_j>r_{j+1}$. The cells $x_j,x_{j+1}$ are incomparable in the cell partial order: if they were comparable, both $T$ and $R$ would have to put them in the same order. Interchanging their consecutive current labels is therefore admissible. It removes exactly one inversion of a permutation from the list of desired labels. Repeating ends at $R$ after exactly the original inversion count, which is the Coxeter length $\ell (\pi )$. ThusThis proves the reduced adjacent-swap path between linear extensions and applies equally to ordinary tableaux.