Suppose we seek to show that
$$v \sum_{k=0}^{n-1} {2n\choose k} \frac{n-k}{n} (v-1)^k
= \sum_{k=0}^{n} {2n\choose k} \frac{2n-2k+1}{2n-k+1} (v-1)^k.$$
Now the coefficient on $[v^q]$ on the left is
$$\sum_{k=0}^{n-1} {2n\choose k} \frac{n-k}{n}
{k\choose q-1} (-1)^{k-(q-1)}$$
and on the right it is
$$\sum_{k=0}^{n} {2n\choose k} \frac{2n-2k+1}{2n-k+1}
{k\choose q} (-1)^{k-q}.$$
Re-factor these to obtain
$$\sum_{k=0}^{n-1} {2n\choose q-1} \frac{n-k}{n}
{2n-(q-1)\choose k-(q-1)} (-1)^{k-(q-1)}
\\ = {2n\choose q-1} \sum_{k=0}^{n-1} \frac{n-k}{n}
{2n-(q-1)\choose k-(q-1)} (-1)^{k-(q-1)} $$
and
$$\sum_{k=0}^{n} {2n\choose q} \frac{2n-2k+1}{2n-k+1}
{2n-q\choose k-q} (-1)^{k-q}
\\ = {2n\choose q} \sum_{k=0}^{n} \frac{2n-2k+1}{2n-k+1}
{2n-q\choose k-q} (-1)^{k-q} .$$
Now the left side can be re-written as follows:
$$\frac{q}{2n-q+1} {2n\choose q} \sum_{k=0}^{n-1} \frac{n-k}{n}
{2n-(q-1)\choose k-(q-1)} (-1)^{k-(q-1)}.$$
We have reduced the problem to the following claim:
$$\frac{q}{2n-q+1} \sum_{k=0}^{n-1} \frac{n-k}{n}
{2n-q+1\choose k-q+1} (-1)^{k-(q-1)}
\\ = \sum_{k=0}^{n} \frac{2n-2k+1}{2n-k+1}
{2n-q\choose k-q} (-1)^{k-q}$$
which is
$$\frac{1}{n} \frac{q}{2n-q+1} \sum_{k=0}^{n-1} (n-k)
{2n-q+1\choose k-q+1} (-1)^{k-(q-1)}
\\ = \frac{1}{2n-2q+1} \sum_{k=0}^{n} (2n-2k+1)
{2n-q+1\choose k-q} (-1)^{k-q}$$
or
$$\frac{q}{n} \sum_{k=0}^{n-1} (n-k)
{2n-q+1\choose k-q+1} (-1)^{k-(q-1)}
\\ = \sum_{k=0}^{n} (2n-2k+1)
{2n-q+1\choose k-q} (-1)^{k-q}.$$
The left here is
$$\frac{q}{n} \sum_{k=1}^{n} k
{2n-q+1\choose n-k-q+1} (-1)^{n-k-(q-1)}
\\ = (-1)^{n-q+1} \frac{q}{n} \sum_{k=0}^{n} k
{2n-q+1\choose n-k-q+1} (-1)^{k} .$$
Introduce
$${2n+1-q\choose n-q-k+1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-q-k+2}} (1+z)^{2n+1-q} \; dz.$$
Observe that this controls the range being zero when $k\gt n-q$ so
we may extend $k$ to infinity to obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-q+2}} (1+z)^{2n+1-q}
(-1)^{n-q+1} \frac{q}{n} \sum_{k\ge 0} (-1)^k k z^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-q+2}} (1+z)^{2n+1-q}
(-1)^{n-q} \frac{q}{n} \frac{z}{(1+z)^2}
\; dz
\\ = (-1)^{n-q} \frac{q}{n}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-q+1}} (1+z)^{2n-1-q}
\; dz
\\ = (-1)^{n-q} \frac{q}{n}
{2n-1-q\choose n-q}
= (-1)^{n-q} \frac{q}{n} {2n-1-q\choose n-1}.$$
For the right we get two pieces call them $R_1$ and $R_2$
which are
$$\sum_{k=0}^{n}
{2n-q+1\choose k-q} (-1)^{k-q}
\\ = (-1)^{n-q} \sum_{k=0}^{n}
{2n-q+1\choose n-k-q} (-1)^{k} $$
and
$$\sum_{k=0}^{n} (2n-2k)
{2n-q+1\choose k-q} (-1)^{k-q}
\\ = 2 (-1)^{n-q} \sum_{k=0}^{n} k
{2n-q+1\choose n-k-q} (-1)^{k}.$$
Introduce once more
$${2n+1-q\choose n-q-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-q-k+1}} (1+z)^{2n+1-q} \; dz$$
to get for $R_1$
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-q+1}} (1+z)^{2n+1-q}
(-1)^{n-q} \sum_{k\ge 0} (-1)^{k} z^k
\; dz
\\ = (-1)^{n-q} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-q+1}} (1+z)^{2n+1-q}
\frac{1}{1+z}
\; dz
\\ = (-1)^{n-q} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-q+1}} (1+z)^{2n-q}
\; dz
\\ = (-1)^{n-q} {2n-q\choose n-q}
= (-1)^{n-q} {2n-q\choose n}.$$
and for $R_2$
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-q+1}} (1+z)^{2n+1-q}
2 (-1)^{n-q} \sum_{k\ge 0} (-1)^{k} k z^k
\; dz
\\ = 2 (-1)^{n-q+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-q+1}} (1+z)^{2n+1-q}
\frac{z}{(1+z)^2}
\; dz
\\ = 2 (-1)^{n-q+1} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-q}} (1+z)^{2n-1-q}
\; dz
\\ = 2 (-1)^{n-q+1} {2n-1-q\choose n-q-1}
= 2 (-1)^{n-q+1} {2n-1-q\choose n}.$$
We have the desired result if we can show that
$$\frac{q}{n} {2n-1-q\choose n-1}
= {2n-q\choose n} - 2{2n-1-q\choose n}.$$
The right is
$$\frac{2n-q}{n} {2n-1-q\choose n-1}
- \frac{2(n-q)}{n} {2n-1-q\choose n-1}.$$
The claim now follows by inspection.
