How it can be shown that:
$$\sum_{k=0}^{n}\binom{2k}{k}\binom{2(n-k)}{n-k}\frac{1}{(2k+1)(2n-2k+1)}=\frac{2^{\left(4n+1\right)}}{\left(n+1\right)^{2}\binom{2n+2}{n+1}}$$
Which is indeed Bruckman’s Formula Version 2.
It seems that there exist a relation between this expression and Rothe–Hagen identity, but I'm not sure.(it's appreciated if someone proves the expression in the most elementary ways, without using hypergeometric functions and integrals)
An elementary approach is WZ-pair. Let $$F(n,k)=\binom{2k}{k}\binom{2(n-k)}{n-k}\binom{2(n+1)}{n+1}\frac{(n+1)^2}{16^n(2k+1)(2(n-k)+1)}$$ Then we have to show that for any $n\geq 0$, $$\sum_{k=0}^{n}F(n,k)=1.$$ Verify that $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k) \tag{1}$$ where $$G(n,k)=-F(n,k)\cdot \frac{k(2k+1)(3n-2k+4)(2(n-k)+1)^2}{4(2(n-k)+3)(n+1)^3(n-k+1)}.$$ The identity (1) seems terrible but it is not. No summation is involved. Moreover $$\binom{2(k+1)}{k+1}=2\binom{2k+1}{k}=\binom{2k+1}{k+1}=\frac{2k+1}{k+1}\binom{2k}{k}$$ implies that all the binomial coefficients in (1) can be easily simplified.
It follows that $$\begin{align}\sum_{k=0}^{n+1}F(n+1,k)-\sum_{k=0}^{n}F(n,k)&=\sum_{k=0}^{n+1} (G(n,k+1)-G(n,k))\\ &=G(n,n+1)-G(n,0)=0 \end{align}$$ because the sum on the right is telescopic. Hence we may conclude that $$\sum_{k=0}^{n}F(n,k)=\sum_{k=0}^{0}F(0,k)=F(0,0)=1.$$
Another approach (less elementary). Note that the Taylor series of arcsin at $0$ is $$\arcsin (x)= \sum_{n=0}^\infty \binom{2k}{k}\frac{x^{2k+1}}{2^{2k}(2k+1)}.$$ The given identity is related to the generating function of $(\arcsin (x))^2$ (see Cauchy product).