We start with the left-hand side. Recall that the value of a cut $S$ is the sum of weights of all the “crossing edges”, meaning, edges which have endpoints in both parts of the partition. Then, combined with the second property of the tree, and the fact that $w_{T}\left(st\right)=w_{T}\left(S_{T|s}\right)$ for any edge $st\in E_{T}$ , we get $$\begin{aligned} w_{T}\left(S\right) & =\sum_{st\in E_{T},s\in S,t\notin S}w_{T}\left(st\right)\\ & =\sum_{st\in E_{T},s\in S,t\notin S}w_{G}\left(S_{T|s}\right) \end{aligned}$$ so it is suffice to show that any crossing edge in the sum of the original graph $w_{G}\left(S\right)$ , contributes also to a sum of $w_{G}\left(S_{T|s}\right)$ for some tree-edge $st\in E_{T}$ such that $s\in S,t\notin S$ . So fix some crossing edge $uv\in E_{G}$ ($u\in S,v\in V\backslash S$ ) from the original graph, and consider the unique path connecting $u$ and $v$ over the tree. Since we starting the path with a vertex from $S$ and ending with a vertex in $V\backslash S$ , there must be an edge $st\in E_{T}$ along the path such that $s\in S$ and $t\in V\backslash S$ . That edge $st$ separates $u$ from $v$ in the tree, so $u\in S_{T|s}$ and $v\notin S_{T|s}$ . Since $uv\in E_{G}$ then it contributes to $w_{G}\left(S_{T|s}\right)$ , which is what we wanted to prove.
We proceed to show the second inequality. For each $st\in E_{T}$ in the sum of $w_{T}\left(S\right)$ , we know that $w_{T}\left(st\right)\le w_{G}\left(S\right)$ since the weight of the tree-edge $w_{T}\left(st\right)$ is equal to the min $st$ -cut in $G$ , thus in particular smaller or equal to the value of the cut $S$ (which also separates $s$ from $t$ but might not be of minimum weight). The inequality follows by summing over all the crossing edges in $S$ over $T$ , which is at most $n-1$ edges.