The lhs and rhs in the first formula differ on $-\frac{4 \sqrt{\pi } \Gamma (n+2)}{3 \Gamma \left(n+\frac{1}{2}\right)}$. For example, then $n=0$ we have $lhs-rhs=-4/3$. But it can be mended.
A natural way to approach this sort of equalities are generating functions. Let
$$ f(x)=\sum_{n=0}^\infty \frac{(4x)^n}{B_n}= \frac{1}{1-x}+\frac{\sqrt{x} \sin ^{-1}\left(\sqrt{x}\right)}{(1-x)^{3/2}},
$$
$$ A_n=\sum_{k=0}^n \frac{4^k}{B_k}. $$
Then
$$ \sum_{n=0}^\infty A_n x^n=\frac{f(x)}{1-x}, $$
$$ f'(x)=\sum_{n=0}^\infty \frac{4^{n+1}x^n}{B_{n+1}},$$
frac{4^{n+1}(n+1)x^n}{B_{n+1}},$$
$$\sum_{n=0}^\infty
\frac{4 \sqrt{\pi } \Gamma (n+2)}{3 \Gamma \left(n+\frac{1}{2}\right)}x^n=
((2 x+(6 \sqrt{x} sin^{-1}(\sqrt{x}))/\sqrt{1-x}+4)/(3 (x-1)^2)).$$
And we indeed have
$$ \frac{f(x)}{1-x}- 2f'(x)+\left(\frac{f(x)}{x}-\frac{1}{x}\right)-\frac1{3(1-x)}=-((2 x+(6 \sqrt{x} sin^{-1}(\sqrt{x}))/\sqrt{1-x}+4)/(3 (x-1)^2)). $$
I think it is possible to obtain formulas for $\sum_{k=0}^n \frac{4^k k^m}{B_k}$, $m\in\mathbb N$ analogously.
