We have
$$
\int_{0}^{1}K\left ( \sqrt{1-x^2} \right )
K\left ( \sqrt{1-x^4} \right )\text{d}x
=\frac{1}{16}{\Gamma\left ( \frac14 \right )^4}
{}_4F_3\left ( \frac14,\frac14,\frac14,\frac14;
\frac12,\frac12,1;1 \right )-\frac14{\Gamma\left ( \frac34 \right )^4}
{}_4F_3\left ( \frac34,\frac34,\frac34,\frac34;
1,\frac32,\frac32;1 \right ).
$$
I am still curious about other approachs. We have it as a manageable series expansion, but seem less dot-able to transform it into hypergeometric series.
Substituting $k^2\rightarrow2/u$ to have the following:
$$
\begin{aligned}
I=\int_{0}^{1}K\left ( \sqrt{1-k^2} \right )
K\left ( \sqrt{1-k^4} \right )\text{d}k
&=\frac{\sqrt{2}}4\int_{2}^{\infty}
\frac{K\left ( \sqrt{1-\frac2{u}} \right ) }{\sqrt{u} }
\frac{2}{u}K\left ( \sqrt{1-\frac{4}{u^2} } \right ) \text{d}u.
\end{aligned}
$$
Noticing that for $u>2$,
$$
\frac{2}{u}K\left ( \sqrt{1-\frac{4}{u^2} } \right )
=\int_1^{u-1}\frac{1}{\sqrt{v^2-1}\sqrt{(u-v)^2-1} }
\text{d}v.
$$
One get
$$
\begin{aligned}
I & =
\frac{\sqrt{2} }{4}
\int_{2}^{\infty} \int_{1}^{u-1}
\frac{K\left ( \sqrt{1-\frac2{u}} \right ) }{\sqrt{u} }
\frac{1}{\sqrt{v^2-1}\sqrt{(u-v)^2-1} }\text{d}v\text{d}u\\
& =
\frac{\sqrt{2} }{4}
\int_{1}^{\infty} \int_{1}^{\infty}
\frac{K\left ( \sqrt{1-\frac2{x+y}} \right ) }{\sqrt{x+y} }
\frac{1}{\sqrt{x^2-1}\sqrt{y^2-1} }\text{d}x\text{d}y.
\end{aligned}
$$
We also notice that for $a\ge1$,
$$
\int_{1}^{\infty} \frac{1}{\sqrt{x^2-1}\sqrt{a+x} }
\text{d}x
=\frac{2}{\sqrt{a+1} } K\left ( \sqrt{\frac{a-1}{a+1} } \right ),
$$
namely,
$$
\int_{1}^{\infty} \frac{1}{\sqrt{z^2-1}\sqrt{x+y+z-1} }
\text{d}z
=\frac{2}{\sqrt{x+y} } K\left ( \sqrt{\frac{x+y-2}{x+y} } \right ).
$$
Therefore,
$$
\begin{aligned}
I&=\frac{\sqrt{2} }{8}
\int_{1}^{\infty} \int_{1}^{\infty}\int_{1}^{\infty}
\frac{1}{\sqrt{x^2-1}\sqrt{y^2-1} \sqrt{z^2-1}}
\frac{1}{\sqrt{x+y+z-1} } \text{d}x
\text{d}y\text{d}z,\\
&=\frac{\sqrt{2} }{8}
\int_{1}^{\infty} \int_{1}^{\infty}\int_{1}^{\infty}
\frac{1}{\sqrt{x^2-1}\sqrt{y^2-1} \sqrt{z^2-1}}
\frac{1}{\sqrt{\pi} } \int_{0}^{\infty}
\frac{e^{-(x+y+z-1)t}}{\sqrt{t} }\text{d}t \text{d}x
\text{d}y\text{d}z,\\
&=\frac{\sqrt{2} }{8\sqrt{\pi} }
\int_{0}^{\infty} \frac{e^x}{\sqrt{x} }
K_0(x)^3\text{d}x,
\end{aligned}
$$
where $K_0(z)$ represents the modified Bessel function of the second kind with order zero and we use its integral representation:
$$
K_0(x)=\int_{1}^{\infty} \frac{e^{-tx}}{\sqrt{t^2-1} }
\text{d}t.
$$
We will also use the first kind,
$$
I_0(x)=\frac{1}{\pi} \int_{-1}^{1}
\frac{e^{-tx}}{\sqrt{1-t^2} }\text{d}t,
$$
which is an even function.
Next we integrate
$$
f(z)=\frac{e^{-z}}{\sqrt{z} }
K_0(z)\left [ K_0(z)+\pi iI_0(z) \right ]^2
$$
along the real-axis, having that the residue at $\infty$ is 0. We obtain $$
\int_{0}^{\infty} \frac{e^{-z}}{\sqrt{z} }
K_0(z)\left [ K_0(z)+\pi iI_0(z) \right ]^2\text{d}z
+\int_{0}^{\infty} \frac{e^{z}}{i\sqrt{z} }
K_0(-z)\left [ K_0(-z)+\pi iI_0(z) \right ]^2\text{d}z=0.
$$
And utilize the analytic continuation $K_0(-z)=K_0(z)-\pi iI_0(z)$ of the upper branch, take the imaginary part and it leaves
$$
\int_{0}^{\infty} \frac{e^z}{\sqrt{z} }
K_0(z)^3\text{d}z
=2\pi\int_{0}^{\infty} \frac{e^{-z}}{\sqrt{z} }
K_0(z)^2I_0(z)
\text{d}z.
$$
This way our desired integral equals to
$$
\begin{aligned}
I=&\frac{\sqrt{2} }{8\sqrt{\pi} }
\int_{0}^{\infty} \frac{e^z}{\sqrt{z} }
K_0(z)^3\text{d}z\\
=&\frac{\sqrt{2}}{4\sqrt{\pi} }
\int_{0}^{\infty} \frac{e^{-z}}{\sqrt{z} }
K_0(z)^2\cdot\pi I_0(z)\text{d}z\\
=&\frac{\sqrt{2} }{4}
\int_{-1}^{1} \int_{1}^{\infty} \int_{1}^{\infty}
\frac{1}{\sqrt{x^2-1}\sqrt{y^2-1}\sqrt{1-z^2} }
\frac{1}{\sqrt{1+x+y+z} } \text{d}x\text{d}y\text{d}z\\
=&\frac{\sqrt{2} }{4}
\int_{1}^{\infty} \int_{1}^{\infty}
\frac{2}{\sqrt{2+x+y} } K\left ( \sqrt{\frac{2}{2+x+y} } \right )
\frac1{\sqrt{x^2-1}\sqrt{y^2-1}}\text{d}x\text{d}y\\
=&\frac{\sqrt{2} }{4}
\int_{2}^{\infty} \int_{1}^{u-1}
\frac{2}{\sqrt{2+u} } K\left ( \sqrt{\frac{2}{2+u} } \right )
\frac1{\sqrt{(u-v)^2-1}\sqrt{v^2-1}}\text{d}v\text{d}u\\
=&\frac{\sqrt{2} }{2}
\int_{2}^{\infty} \frac{K\left ( \sqrt{\frac{2}{2+u}}\right) }{\sqrt{2+u} }
\frac{2}{u}K\left ( \sqrt{1-\frac{4}{u^2} } \right )\text{d}u\\
=&2\int_{0}^{1} \frac{K\left ( \frac{k}{\sqrt{1+k^2} } \right ) K\left ( \sqrt{1-k^4} \right ) }{\sqrt{1+k^2} } \text{d}k
\\
=&\pi \sum_{n=0}^{\infty} \left ( -1 \right )^n \frac{\left ( \frac12 \right )_n^2 }{(1)_n^2}
\int_{0}^{1}k^{2n} K\left ( \sqrt{1-k^4} \right ) \text{d}k\\
=&\pi \sum_{n=0}^{\infty} \left ( -1 \right )^n \frac{\left ( \frac12 \right )_n^2 }{(1)_n^2}
\cdot\frac{\pi}{8} \frac{\Gamma\left ( \frac{n}{2}+\frac14 \right )^2 }{\Gamma\left ( \frac{n}{2}+\frac34 \right )^2}\\
=&\frac{1}{16}{\Gamma\left ( \frac14 \right )^4}
{}_4F_3\left ( \frac14,\frac14,\frac14,\frac14;
\frac12,\frac12,1;1 \right )-\frac14{\Gamma\left ( \frac34 \right )^4}
{}_4F_3\left ( \frac34,\frac34,\frac34,\frac34;
1,\frac32,\frac32;1 \right ).
\end{aligned}
$$
As to a generalization, we have
$$
\int_{0}^{1} k^2K\left (\sqrt{ 1-k^4 }\right )
\left [ \frac{K\left ( \sqrt{1-k^2} \right )
-E\left ( \sqrt{1-k^2} \right ) }{1-k^2}
+\frac{2}{\sqrt{1+k^2} } E\left ( \frac{k}{\sqrt{1+k^2} } \right ) \right ]
\text{d} k=\frac{\pi^2}{4}.
$$
