Update. OP asked for a complete proof and that is
indeed what we will present.
Supposing we start from
$$\sum_{k=0}^n {n\choose k} {n+r\choose k} {2n\choose 2k}^{-1}.$$
Write
$${n+r\choose k} = \frac{(n+r)!}{k! \times (n+r-k)!}
= {n+r\choose n} \frac{n! \times r!}{k! \times (n+r-k)!}
\\ = {n+r\choose n} {n\choose k} {n-k+r\choose r}^{-1}.$$
Furthermore
$${n\choose k}^2 {2n\choose 2k}^{-1}
= {2n\choose n}^{-1}
\frac{(2k)! \times (2n-2k)!}{(n-k)!^2\times k!^2}
\\ = {2n\choose n}^{-1} {2k\choose k} {2n-2k\choose n-k}.$$
We thus have for our sum
$${n+r\choose n} {2n\choose n}^{-1}
\sum_{k=0}^n {2k\choose k} {2n-2k\choose n-k}
{n-k+r\choose r}^{-1}.$$
Recall from MSE
4316307 the
following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= [v^n] \log\frac{1}{1-v} (v-1)^{n-k}.$$
We can re-write this as
$${n-1\choose k-1}^{-1} =
n [v^n] \log\frac{1}{1-v} (v-1)^{n-k}.$$
We have for the sum without the scalar in front
$$\sum_{k=0}^n {2k\choose k} {2n-2k\choose n-k}
(n+1-k+r) [v^{n+1-k+r}] \log \frac{1}{1-v}
(v-1)^{n-k}.$$
The contribution from $v$ is
$$\;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{n+2-k+r}}
\log\frac{1}{1-v} (v-1)^{n-k}.$$
Now put $v/(1-v)=z$ so that $v=z/(1+z)$ and $dv= 1/(1+z)^2 \; dz$
to obtain
$$\;\underset{z}{\mathrm{res}}\;
\frac{1}{z^{n+2-k+r}} (-1)^{n-k} (1+z)^{r+2}
\log (1+z) \frac{1}{(1+z)^2}.$$
We thus find for the sum
$$ (-1)^n [z^{n+1+r}] \log(1+z) (1+z)^r
\\ \times \sum_{k=0}^n {2k\choose k} {2n-2k\choose n-k}
(n+1-k+r) (-1)^k z^k
\\ = (-1)^r [z^{n+1+r}] \log\frac{1}{1-z} (1-z)^r
\\ \times \sum_{k=0}^n {2k\choose k} {2n-2k\choose n-k}
(n+1-k+r) z^k
\\ = [z^{n+1+r}] \log\frac{1}{1-z} (z-1)^r
\\ \times \sum_{k=0}^n {2k\choose k} {2n-2k\choose n-k}
(n+1-k+r) z^k.$$
We now get two pieces.
First piece
This is
$$(n+1+r) [z^{n+1+r}] \log\frac{1}{1-z} (z-1)^r
\\ \times \sum_{k=0}^n {2k\choose k} {2n-2k\choose n-k} z^k
\\ = (n+1+r) [z^{n+1+r}] \log\frac{1}{1-z} (z-1)^r
[w^n] \frac{1}{\sqrt{1-4wz}} \frac{1}{\sqrt{1-4w}}.$$
The square root yields
$$[w^n] \frac{1}{\sqrt{1-4w-4w(z-1)}} \frac{1}{\sqrt{1-4w}}
\\ = [w^n] \frac{1}{\sqrt{1-4w(z-1)/(1-4w)}} \frac{1}{1-4w}
\\ = [w^n] \sum_{p=0}^n
{2p\choose p} \frac{w^p (z-1)^p}{(1-4w)^{p+1}}
= \sum_{p=0}^n
{2p\choose p} (z-1)^p 4^{n-p} {n\choose p}.$$
Applying the logarithm
$$\sum_{p=0}^n
{2p\choose p} 4^{n-p} {n\choose p}
{n+r\choose n-p}^{-1}.$$
Next observe that
$${n\choose p} {n+r\choose n-p}^{-1}
= \frac{n! \times (r+p)!}{p! \times (n+r)!}
\\ = {n+r\choose n}^{-1} {r+p\choose r}$$
which produces
$${n+r\choose n}^{-1}
\sum_{p=0}^n {2p\choose p} 4^{n-p} {r+p\choose r}.$$
Second piece
This is
$$- [z^{n+1+r}] \log\frac{1}{1-z} (z-1)^r
\\ \times \sum_{k=0}^n k {2k\choose k} {2n-2k\choose n-k} z^k
\\ = - [z^{n+1+r}] \log\frac{1}{1-z} (z-1)^r
[w^n] \frac{2wz}{\sqrt{1-4wz}^3} \frac{1}{\sqrt{1-4w}}.$$
The square root yields
$$[w^n] \frac{2wz}{\sqrt{1-4w-4w(z-1)}^3} \frac{1}{\sqrt{1-4w}}
\\ = z [w^{n}] \frac{2w}{\sqrt{1-4w(z-1)/(1-4w)}^3}
\frac{1}{(1-4w)^2}
\\ = \frac{z}{z-1} [w^{n}] \frac{2w(z-1)/(1-4w)}
{\sqrt{1-4w(z-1)/(1-4w)}^3}
\frac{1}{1-4w}
\\ = \frac{z}{z-1} [w^{n}] \sum_{p=0}^n
p {2p\choose p} \frac{w^p (z-1)^p}{(1-4w)^{p+1}}
= \frac{z}{z-1} \sum_{p=0}^n
p {2p\choose p} (z-1)^p 4^{n-p} {n\choose p}.$$
Applying the logarithm and the sign
$$- \frac{1}{n+r} \sum_{p=0}^n
p {2p\choose p} 4^{n-p} {n\choose p}
{n+r-1\choose n-p}^{-1}.$$
Next observe that
$$\frac{1}{n+r} {n\choose p} {n+r-1\choose n-p}^{-1}
= \frac{n! \times (r+p-1)!}{p! \times (n+r)!}
\\ = \frac{1}{r+p} {n+r\choose n}^{-1} {r+p\choose r}$$
which produces
$$- {n+r\choose n}^{-1}
\sum_{p=0}^n \frac{p}{r+p}
{2p\choose p} 4^{n-p} {r+p\choose r}.$$
Join the two pieces
We join the two pieces and activate the scalars to get
$${2n\choose n}^{-1}
\sum_{p=0}^n \frac{r}{r+p}
{2p\choose p} 4^{n-p} {r+p\choose r}.$$
For this to hold we need $r\ge 1.$ We further obtain
$${2n\choose n}^{-1}
\sum_{p=0}^n
{2p\choose p} 4^{n-p} {r+p-1\choose r-1}.$$
Working with the sum
$$4^n [z^n] \frac{1}{1-z} \sum_{p\ge 0} z^p
{2p\choose p} 4^{-p} {r+p-1\choose r-1}
\\ = 4^n [z^n] \frac{1}{1-z}
[w^{r-1}] (1+w)^{r-1} \sum_{p\ge 0} z^p
{2p\choose p} 4^{-p} (1+w)^p
\\ = 4^n [z^n] \frac{1}{1-z}
[w^{r-1}] (1+w)^{r-1}
\frac{1}{\sqrt{1-z(1+w)}}
\\ = 4^n [z^n] \frac{1}{(1-z)^{3/2}}
[w^{r-1}] (1+w)^{r-1}
\frac{1}{\sqrt{1-wz/(1-z)}}
\\ = 4^n [z^n] \frac{1}{(1-z)^{3/2}}
\sum_{p=0}^{r-1} {r-1\choose r-1-p} {2p\choose p}
4^{-p} \frac{z^p}{(1-z)^p}
\\ = 4^n
\sum_{p=0}^{r-1} {r-1\choose p} {2p\choose p}
4^{-p} {n+1/2\choose n-p}.$$
The last binomial coefficient is zero for a negative lower index by
construction. We have for $r\ge 1$ the closed form
$$\bbox[5px,border:2px solid #00A000]{
4^n {2n\choose n}^{-1}
\sum_{p=0}^{r-1} {r-1\choose p} {2p\choose p}
4^{-p} {n+1/2\choose n-p}.}$$
This gives e.g. for $r=1$
$$4^n {2n\choose n}^{-1} {n+1/2\choose n}.$$
We get for $r=2$
$$4^n {2n\choose n}^{-1}
\left[ {n+1/2\choose n} + \frac{1}{2} {n+1/2\choose n-1} \right].$$
One more example is $r=3$ which yields
$$4^n {2n\choose n}^{-1}
\left[ {n+1/2\choose n} + {n+1/2\choose n-1}
+ \frac{3}{8} {n+1/2\choose n-2} \right].$$
Last example is $r=4$
$$4^n {2n\choose n}^{-1}
\left[ {n+1/2\choose n} + \frac{3}{2} {n+1/2\choose n-1}
+ \frac{9}{8} {n+1/2\choose n-2}
+ \frac{5}{16} {n+1/2\choose n-3} \right].$$
The case of $r=0$
We have from the introduction
$${n+r\choose n} {2n\choose n}^{-1}
\sum_{k=0}^n {2k\choose k} {2n-2k\choose n-k}
{n-k+r\choose r}^{-1}.$$
Evaluate at $r=0$ to get
$${2n\choose n}^{-1}
\sum_{k=0}^n {2k\choose k} {2n-2k\choose n-k}
\\ = {2n\choose n}^{-1}
[z^n] \frac{1}{\sqrt{1-4z}} \frac{1}{\sqrt{1-4z}}
= {2n\choose n}^{-1}
[z^n] \frac{1}{1-4z}
= 4^n {2n\choose n}^{-1}.$$
