$$ \begin{align} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}\frac{4^k}{{2k \choose k}} &= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{2k+1} \int_{0}^{\pi /2} \sin^{2k+1} (x) \, dx \\ &= \int_{0}^{\pi /2} \sum_{n=0}^{\infty} \frac{(-1)^{k} \sin^{2k+1} (x)}{2k+1} \, dx\\ &= \int_{0}^{\pi /2} \arctan (\sin x) \, dx \\ &= \int_{0}^{1} \frac{\arctan t}{\sqrt{1-t^{2}}} \, dt \end{align}$$
Let $ \displaystyle I(a) = \int_{0}^{1} \frac{\arctan (at)}{\sqrt{1-t^{2}}} \ dt$.
Then differentiating under the integral sign,
$$ \begin{align} I'(a) &= \int_{0}^{1} \frac{t}{(1+a^{2}t^{2})\sqrt{1-t^{2}}} \, dt \\ &= \int_{0}^{1} \frac{1}{[1+a^{2}(1-u^{2})]u} \, u \, du \\ &= \frac{1}{1+a^{2}} \int_{0}^{1} \frac{1}{1-\left( \frac{au}{\sqrt{1+a^{2}}}\right)^{2}} \, du \\ &= \frac{1}{a \sqrt{1+a^{2}}} \text{arctanh} \left( \frac{a}{\sqrt{1+a^{2}}} \right) \\ &= \frac{1}{a\sqrt{1+a^{2}}} \frac{1}{2} \ln \Big((a+\sqrt{1+a^{2}})^{2} \Big) \\ &= \frac{1}{a \sqrt{1+a^{2}}} \ln \left( a+ \sqrt{1+a^{2}} \right) \\ &= \frac{1}{a \sqrt{1+a^{2}}} \text{arcsinh}(a) . \end{align}$$
And then integrating back,
$$ \begin{align} I(1)-I(0) = I(1) &= \int_{0}^{1} \frac{\text{arcsinh}(a)}{a \sqrt{1+a^{2}}} \, da \\ &= - \text{arcsinh}(a) \text{arcsinh}(\frac{1}{a}) \Bigg|^{1}_{0} + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \, da \\ &= - \text{arcsinh}^{2}(1) + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \, da \\ &= - \ln^{2}(1+\sqrt{2}) + \int_{0}^{1} \frac{\text{arcsinh}(\frac{1}{a})}{\sqrt{1+a^{2}}} \, da . \end{align}$$
Now let $ \displaystyle w = \frac{1}{a}$.
Then
$$ I(1) = - \ln^{2}(1+\sqrt{2}) + \int_{1}^{\infty} \frac{\text{arcsinh}(w)}{w \sqrt{1+w^{2}}} \, dw$$
$$ = - \ln^{2}(1+\sqrt{2}) + I(\infty) - I(1) .$$
Therefore,
$$ \begin{align} I(1) &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{I(\infty)}{2} \\ &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi}{4} \int_{0}^{1} \frac{1}{\sqrt{1-t^{2}}} \, dt \\ &= - \frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi^{2}}{8} . \end{align}$$
Let ${n \brace k}$ denote the $(n,k)^{th}$ Stirling number of the second kind. Using the well-known identity
$$
k^m=\sum_{j=0}^m {m \brace j}\frac{k!}{(k-j)!},
$$
we have
$$
\begin{align}
\sum_{k=0}^n \binom{n}k k^m x^k y^{n-k}
&= \sum_{k=0}^n \binom{n}k x^k y^{n-k}\sum_{j=1}^m {m\brace j}\frac{k!}{(k-j)!}
\\&= \sum_{j=1}^m\frac{n!}{(n-j)!}{m \brace j}\sum_{k=0}^n \binom{n-j}{k-j}x^ky^{n-k}
\\&= \sum_{j=1}^m \frac{n!}{(n-j)!}{m \brace j}x^j(x+y)^{n-j}
\end{align}
$$
In the special case where $y=1-x$ that you mentioned, this simplifies a bit more to
$$
\sum_{j=1}^m \frac{n!}{(n-j)!}{m \brace j}x^j
$$
In this special case, the summation is also equal to the $m^{th}$ binomial moment. That is, it is the expected value of $X^m$ when $X$ has the $\text{Binomial}(n,x)$ distribution. There's some more info about these moments on Wikipedia, including an upper bound, but no lower bound.
Best Answer
Absorb the $k+1$ and apply Vandermonde: \begin{align} \sum_{k=0}^n \frac{\binom{n}{k}^2}{k+1} &=\frac{1}{n+1}\sum_{k=0}^n \frac{n+1}{k+1}\binom{n}{k}^2\\ &=\frac{1}{n+1}\sum_{k=0}^n \binom{n+1}{k+1}\binom{n}{k}\\ &=\frac{1}{n+1}\sum_{k=0}^n \binom{n+1}{n-k}\binom{n}{k}\\ &=\frac{1}{n+1}\binom{2n+1}{n} \end{align}