Let's re-index the sum on the LHS of the problem (for the below convenience) so that
$$\frac1s\sum_{k=0}^{s-1}\binom{s+k-1}k(s-k)v^{s-k}(v-1)^k
=\frac{v}s\sum_{k=0}^{s-1}\binom{2s}k(s-k)(v-1)^k.$$
Or, just pull out the $v$ factor (leaving behind $\frac1s$, again for convenience):
$$\frac1s\sum_{k=0}^{s-1}\binom{s+k-1}k(s-k)v^{s-1-k}(v-1)^k
=\frac1s\sum_{k=0}^{s-1}\binom{2s}k(s-k)(v-1)^k.\tag1$$
We apply an automated technique called the Wilf-Zeilberger method for which Zeilberger wrote an algorithm, now implemented in symbolic softwares such as Maple and Mathematica.
To prove the above identity, we show both sides of (1) satisfy the same non-homogeneous recurrence $x(s+1)-v^2x(s)=\binom{2s}{s+1}\frac{(v-1)^{s+1}}s$ and just check one initial condition, say at $s=1$.
Suppressing other variables, let $F_1(s,k)=\binom{s+k-1}k\frac{s-k}sv^{s-1-k}(v-1)^k$ and $F_2(s,k)=\binom{2s}k\frac{s-k}s(v-1)^k$. The above method generates the WZ-mates
\begin{align} G_1(s,k)&=F_1(s,k)
\frac{v^2(s+2-k)k}{(s-k)(s+1)}, \\
G_2(s,k)&=F_2(s,k)
\frac{(2s^2v+4sv+2s^2+3s-3svk-3kv+2v-2sk+vk^2)k}{(2s-k+1)(2s+2-k)(s-k)}
\end{align}
satisfying the relations (check using a computer)
\begin{align}
F_1(s+1,k)-v^2F_1(s,k)&=G_1(s,k+1)-G_1(s,k), \\
F_2(s+1,k)-v^2F_2(s,k)&=G_2(s,k+1)-G_2(s,k). \tag2
\end{align}
Denote $a(s)=\sum_{k=0}^{s-1}F_1(s,k)$ and $b(s)=\sum_{k=0}^{s-1}F_2(s,k)$.
Now, sum both equations in (2) over integers $0\leq k\leq s$. For instance, the first equation in (2) gives
$$a(s+1)-v^2a(s)-v^2F_1(s,s)=G_1(s,s+1)-G_1(s,0).$$
But, $F_1(s,s)=0=G_1(s,0)$ and hence $a(s+1)-v^2a(s)=G_1(s,s+1)$.
Similarly, the second equation in (2) leads to
$$b(s+1)-v^2b(s)-v^2F_2(s,s)=G_2(s,s+1)-G_2(s,0).$$
But, $F_2(s,s)=0=G_2(s,0)$ and hence $b(s+1)-v^2b(s)=G_2(s,s+1)$.
However, after some algebraic simplification, we obtain
$$G_1(s,s+1)=G_2(s,s+1)=\binom{2s}{s+1}\frac{(v-1)^{s+1}}s.$$
Since $a(1)=b(1)$, it follows that $a(s)=b(s)$. The proof is complete.