$\beta'(-1)$ can actually be found directly from the functional equation. We have
$$ \beta'(-1) = - \frac{\mathrm{d}}{\mathrm{d}s} \left[\left(\frac{2}{\pi}\right)^s \sin \left(\frac{\pi}{2} s\right) \Gamma(s) \beta(s)\right] ~ \Bigg\rvert_{s=2} \, .$$
Now note that the only non-vanishing term on the right-hand side is the one containing the derivative of the sine. Therefore
$$\beta'(-1) = \left(\frac{2}{\pi}\right)^2 \frac{\pi}{2} [-\cos(\pi)] \Gamma(2) \beta(2) = \frac{2 \mathrm{K}}{\pi} \, .$$
However, this also means that $\beta'(2)$ cannot be calculated from $\beta'(-1)$.
$\beta'(1)$ is not as easy to find. We can start from the integral
\begin{align} \frac{1}{2} \int \limits_0^\infty \frac{-\ln(x)}{\cosh(x)} \, \mathrm{d} x &= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}s} \int \limits_0^\infty \frac{x^{s-1}}{\cosh(x)} \, \mathrm{d} x ~ \Bigg\rvert_{s=1} = - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}s} [2 \Gamma(s) \beta(s)] ~ \Bigg\rvert_{s=1} \\
&= \frac{\pi \gamma}{4} - \beta'(1) \, .
\end{align}
Since the left-hand side is a representation of Vardi's integral
$$ \int \limits_0^1 \frac{-\ln(-\ln(x))}{1+x^2} \, \mathrm{d} x = \frac{\pi}{4} \ln\left(\frac{\operatorname{\Gamma}^4\left(\frac{1}{4}\right)}{4\pi^3}\right) \, ,$$
we obtain the desired result, from which the value of $\beta'(0)$ follows by the functional equation.
Various ways to evaluate the integral are discussed in this paper by Blagouchine. By the way, the author convincingly argues that the integral should actually be called Malmsten's integral, since Malmsten first computed it in 1842, which is in line with the claim from the Wikipedia page.
As an alternative route, here is a brief sketch of how I have calculated $\beta'(1)$ :
Plugging $z=\frac{1}{4}$ into Binet's first log-gamma formula, we obtain (after some algebra and integration by parts/substitutions)
\begin{align}
\ln\left(\frac{\operatorname{\Gamma}^4\left(\frac{1}{4}\right)}{4\pi^2}\right) &= 2 \ln(2) - 1 + \int \limits_0^\infty \frac{\tanh(x)}{x} \mathrm{e}^{-x} \, \mathrm{d} x \\ &\phantom{=} + \int \limits_0^\infty \left[\frac{\coth(x)}{x} -\frac{1}{\sinh^2 (x)}\right] \mathrm{e}^{-x} \, \mathrm{d} x - \int \limits_0^\infty \left[\coth(x)-\frac{1}{x}\right] \mathrm{e}^{-x} \, \mathrm{d} x \\
&\equiv 2\ln(2) - 1 + I_1 + I_2 - I_3 \, .
\end{align}
We can then compute
\begin{align}
I_1 &= \gamma + \ln \left(\frac{\pi}{2}\right) - \frac{4}{\pi} \beta'(1) \, ,\\
I_2 &= \gamma \, ,\\
I_3 &= \gamma + \ln(2) - 1 \,
\end{align}
and solve for $\beta'(1)$ .
In order to derive the functional equation for $\beta$ we consider $s \in \mathbb{C}$ with $0 < \operatorname{Re}(s) < 1$ first. Using the geometric series we obtain
$$ \operatorname{\beta} (1 - s) = \frac{1}{2 \operatorname{\Gamma} (1-s)} \int \limits_0^\infty \frac{t^{-s}}{\cosh(t)} \, \mathrm{d} t = \frac{1}{4 \operatorname{\Gamma} (1-s)} \int \limits_{-\infty}^\infty \lvert t \rvert^{-s}\operatorname{sech}(t) \, \mathrm{d} t \, . $$
We have $\mathcal{F} (\lvert \cdot \rvert^{-s}) (k) = 2 \operatorname{\Gamma}(1-s)\sin \left(\frac{\pi}{2} s\right) \lvert k \rvert^{s-1}$ from this question and $\mathcal{F} (\operatorname{sech}) (k) = \pi \operatorname{sech} \left(\frac{\pi}{2} k\right)$ from here, so we can use Parseval's theorem (we are cheating a bit here, since the first function is not in $L^2 (\mathbb{R})$, but this is easily fixed by considering $\lvert t \rvert^{-s} \mathrm{e}^{-\varepsilon \lvert t \rvert}$ instead and taking the limit $\varepsilon \to 0^+$) to conclude that
\begin{align}
\beta(1 - s) &= \frac{1}{8 \pi \operatorname{\Gamma} (1-s)} \int \limits_{-\infty}^\infty 2 \operatorname{\Gamma}(1-s)\sin \left(\frac{\pi}{2} s\right) \lvert k \rvert^{s-1} \pi \operatorname{sech}\left(\frac{\pi}{2} k\right) \, \mathrm{d} k \\
&= \left(\frac{2}{\pi}\right)^s \sin \left(\frac{\pi}{2} s\right) \frac{1}{2} \int \limits_0^\infty \frac{u^{s-1}}{\cosh(u)} \, \mathrm{d} u = \left(\frac{2}{\pi}\right)^s \sin \left(\frac{\pi}{2} s\right) \operatorname{\Gamma} (s) \operatorname{\beta} (s)
\end{align}
holds for $0 < \operatorname{Re} (s) < 1$. Since the right-hand side is holomorphic on the right half-plane we can use this equation to extend $\beta$ to an entire function. Then the functional equation holds on all of $\mathbb{C}$ by the identity theorem.
