I think my process isn't helpful indeed. And I'll outline my original proof here. We deal with it by two claims.
Claim 1.
For a positive real $a\ge1$, we have
$$
\int_{0}^{1} \frac{K(k)}{\sqrt{a-k} }\text{d}k
=\frac{1}{2} \int_{0}^{\frac1a} \frac{K^\prime(k)}{\sqrt{k}\sqrt{1-ak} }
\text{d} k
-\int_{\frac1a}^1 \frac{K(k)}{\sqrt{k}\sqrt{ak-1} }
\text{d} k.
$$
Proof. Recalling the analytic continuation of complete elliptic integral $K(k)=\frac{1}{\left | k \right | } \left ( K\left ( \frac{1}{k} \right )
\pm iK^\prime\left (\frac{1}{k} \right ) \right ),\Re(k)>1$, where the symbol is determined by the branch cuts chosen, we integrate the functions along the real axis respectively,
$$
f(z)=\frac{K(z)}{\sqrt{a+z} }
,\frac{K\left ( z \right ) }{\sqrt{z} \sqrt{1+az} }.
$$
We have
$$
0=\int_{0}^{1} \frac{K(z)}{\sqrt{a+z} }\text{d}z
+\int_{0}^{1} \frac{K(z)}{\sqrt{a-z} }\text{d}z
+\int_{1}^{\infty}
\frac{1}{\sqrt{a+z} } \frac{1}{z} \left ( K\left ( \frac{1}{z} \right )
+i K^\prime\left (\frac{1}{z} \right ) \right )\text{d}z
+\int_{1}^{a}
\frac{1}{\sqrt{a-z} } \frac{1}{z} \left ( K\left ( \frac{1}{z} \right )
- iK^\prime\left (\frac{1}{z} \right ) \right )\text{d}z
+\int_{a}^{\infty}
\frac{1}{i\sqrt{z-a} } \frac{1}{z} \left ( K\left ( \frac{1}{z} \right )
- iK^\prime\left (\frac{1}{z} \right ) \right )\text{d}z,$$
$$
0=\int_{0}^{1} \frac{K(z)}{\sqrt{z}\sqrt{1+az} }\text{d}z
+\int_{1}^{\infty}
\frac{1}{\sqrt{z}\sqrt{1+az} }\frac{1}{z} \left ( K\left ( \frac{1}{z} \right )
+ iK^\prime\left (\frac{1}{z} \right ) \right )\text{d}z
+\int_{0}^{\frac{1}{a} }
\frac{K(z)}{i\sqrt{z}\sqrt{1-az} }\text{d}z
+\int_{\frac{1}{a} }^1
\frac{K(z)}{i\sqrt{z}\cdot i\sqrt{az-1} }\text{d}z
+\int_1^\infty
\frac{1}{i\sqrt{z}\cdot i\sqrt{az-1} }
\frac{1}{z} \left ( K\left ( \frac{1}{z} \right )
- iK^\prime\left (\frac{1}{z} \right ) \right )\text{d}z.
$$
By extracting the real part from above, one obtain
$$
0=\int_{0}^{1} \frac{K(z)}{\sqrt{a+z} } \text{d}z
+\int_{0}^{1} \frac{K(z)}{\sqrt{a-z} } \text{d}z
+\int_{0}^{1} \frac{K(z)}{\sqrt{z}\sqrt{1+az} } \text{d}z
+\int_{\frac1a}^{1} \frac{K(z)}{\sqrt{z}\sqrt{az-1} }\text{d}z
-\int_{0}^{\frac1a} \frac{K^\prime(z)}{\sqrt{z}\sqrt{1-az} }\text{d}z,
$$
$$
0=\int_{0}^{1} \frac{K(z)}{\sqrt{z}\sqrt{1+az} } \text{d}z
+\int_{0}^{1} \frac{K(z)}{\sqrt{a+z} } \text{d}z
-\int_{\frac1a}^{1} \frac{K(z)}{\sqrt{z}\sqrt{az-1} }\text{d}z
-\int_{0}^{1} \frac{K(z)}{\sqrt{a-z} } \text{d}z.
$$
Comparing two equations proves our claim.
Next we substituting $a=3$, Claim 1 allows us to have
$$
\int_{0}^{1} \frac{K(k)}{\sqrt{3-k} }\text{d}k
=\frac{1}{2} \int_{0}^{\frac13} \frac{K^\prime(k)}{\sqrt{k}\sqrt{1-3k} }
\text{d} k
-\int_{\frac13}^1 \frac{K(k)}{\sqrt{k}\sqrt{3k-1} }
\text{d} k.
$$By Landen's transformation, i.e. substituting $k\rightarrow(1-k)/(1+k)$ and $K\left ( \frac{1-k}{1+k} \right )
=\frac{1+k}{2}K^\prime(k),K^\prime\left ( \frac{1-k}{1+k} \right )
=(1+k)K(k),$ one have
$$\begin{aligned}
&\int_{0}^{\frac13} \frac{K^\prime(k)}{\sqrt{k}\sqrt{1-3k} }=\sqrt{2} \int_{\frac12}^{1} \frac{K(k)}{
\sqrt{1-k}\sqrt{2k-1 } } \text{d}k=
2\int_{0}^{1} \frac{1}{\sqrt{1-k^2} }
K\left ( \frac{1+k^2}{2} \right ) \text{d}k,\\
&\int_{\frac13}^1 \frac{K(k)}{\sqrt{k}\sqrt{3k-1} }=\frac1{\sqrt{2}} \int_0^{\frac12} \frac{K^\prime(k)}{
\sqrt{1-k}\sqrt{1-2k } } \text{d}k=
\int_{0}^{1} \frac{1}{\sqrt{1+k^2} }
K^\prime\left ( \frac{1-k^2}{2} \right ) \text{d}k.
\end{aligned}
$$
Then we need the second claim.
Claim 2. Denoting $I=\int_{0}^{\infty} e^{-3x}(\pi I_0(x))^3\text{d}x=\int_{\left [ 0,\pi \right ]^3 }
\frac{1}{3-\cos x-\cos y-\cos z}\text{d}x\text{d}y\text{d}z,$
the following relations holds
$$
\int_{0}^{1} \frac{1}{\sqrt{1-k^2} }
K\left ( \frac{1+k^2}{2} \right ) \text{d}k=\frac{1}{\pi\sqrt{2}}I,\qquad
\int_{0}^{1} \frac{1}{\sqrt{1+k^2} }
K^\prime\left ( \frac{1-k^2}{2} \right ) \text{d}k=\frac{\sqrt{2}}{3\pi}I.
$$
Proof. The first is by direct computation(see @Prezmo's answer). For the second, we start from substituting $\frac{1-k^2}2=\frac2{3+x}$, which gives
$$
\int_{0}^{1} \frac{1}{\sqrt{1+k^2} }
K^\prime\left ( \frac{1-k^2}{2} \right ) \text{d}k
= \sqrt{2} \int_{1}^{\infty}
\frac{1}{\sqrt{x^2-1}\left ( 3+x \right ) }
K^\prime\left ( \frac{2}{3+x} \right )\text{d}x.
$$
Now using the following identity valid for $x>2$,
$$
\int_{1}^{\infty} \frac{1}{\sqrt{t^2-1}
\sqrt{\left ( x+t \right )^2-1 } } \text{d}t
=\frac{1}{x} K^\prime\left ( \frac{2}{x} \right ),
$$
we see that$$
\int_{0}^{1} \frac{1}{\sqrt{1+k^2} }
K^\prime\left ( \frac{1-k^2}{2} \right ) \text{d}k
=\sqrt{2} \int_{1}^{\infty}
\int_{1}^{\infty} \frac{1}{\sqrt{x^2-1}\sqrt{y^2-1} \sqrt{\left ( 3+x+y \right )^2-1} }
\text{d} x\text{d}y.
$$
Empolying a simple integral $\frac{1}{\sqrt{a^2-1} }
=\frac{1}{\pi} \int_{-1}^{1} \frac{1}{\sqrt{1-z^2}
\left ( a+z \right ) }\text{d}z$, we obtain
$$
\begin{aligned}
\int_{0}^{1} \frac{1}{\sqrt{1+k^2} }
K^\prime\left ( \frac{1-k^2}{2} \right ) \text{d}k
& = \sqrt{2} \int_{1}^{\infty}
\int_{1}^{\infty} \frac{1}{\sqrt{x^2-1}\sqrt{y^2-1} \sqrt{\left ( 3+x+y \right )^2-1} }
\text{d} x\text{d}y,\\
&=\frac{\sqrt{2} }{\pi} \int_{1}^{\infty} \int_{1}^{\infty}
\int_{-1}^{1} \frac{1}{\sqrt{1-x^2}\sqrt{y^2-1}\sqrt{z^2-1} } \frac{1}{3+x+y+z}
\text{d}x\text{d}y\text{d}z,\\
&=\frac{\sqrt{2} }{\pi} \int_{0}^{\infty}
e^{-3x}K_0(x)^2\cdot \pi I_0(x) \text{d}x,
\end{aligned}
$$
by invoking modified Bessel functions of the first kind $I_0(x)=\frac1\pi\int_{-1}^{1}
\frac{e^{-xy}}{\sqrt{1-y^2} } \text{d}y$ and second kind $K_0(x)=\int_{1}^{\infty}
\frac{e^{-xy}}{\sqrt{y^2-1} } \text{d}y$ with zero order. Again by recalling the analytic continuation of $K_0(z)$, and integrating $
f(z)=e^{-3z}\left(K_0(z)+\pi i I_0(z)\right)^3$ along the real axis, we have
$$
\begin{aligned}
0=&\int_{0}^{\infty} e^{-3z}\left(K_0(z)+\pi i I_0(z)\right)^3
\text{d}z+
\int_{0}^{\infty} e^{3z}\left(K_0(-z)+\pi i I_0(z)\right)^3\text{d}z\\
=&\int_{0}^{\infty} e^{-3z}\left(K_0(z)+\pi i I_0(z)\right)^3
\text{d}z+
\int_{0}^{\infty} e^{3z}K_0(z)^3\text{d}z.
\end{aligned}
$$
Extract the imaginary part to obtain
$$
\int_{0}^{\infty} e^{-3x}K_0(x)^2\cdot \pi I_0(x)\text{d}x
=\frac{1}{3} \int_{0}^{\infty} e^{-3x}\left ( \pi I_0(x) \right )^3
\text{d} x=\frac13I,
$$
which leads to the Claim 2.
Finally our desired integral is
$$
\begin{aligned}
\int_{0}^{1} \frac{K(k)}{\sqrt{3-k} } \text{d}k
& = \int_{0}^{1} \frac{1}{\sqrt{1-k^2} }
K\left ( \frac{1+k^2}{2} \right ) \text{d}k
-\int_{0}^{1} \frac{1}{\sqrt{1+k^2} }
K^\prime\left ( \frac{1-k^2}{2} \right ) \text{d}k,\\
&=\frac{\sqrt{2} }{6\pi} I,\\
&=\frac{1}{96\pi\sqrt{3} } \Gamma\left ( \frac1{24} \right )
\Gamma\left ( \frac5{24} \right ) \Gamma\left ( \frac7{24} \right )
\Gamma\left ( \frac{11}{24} \right ).
\end{aligned}
$$
The last line is due to the third Watson's triple integral, which is well-known but non-trivial conclusion.
Remark 1. A possible method is by considering differential equations. Define
$$
I(a)=\int_{0}^{1} \frac{K(k)}{\sqrt{a-k} } \text{d}k,\quad a\ge1,
$$
we have
$$
\left ( 8a(a^2-1)\partial_a^3
+12(3a^2-1) \partial_a^2
+26a\partial_a+1\right ) I(a)
=-\frac{4}{\left ( a-1 \right )^{3/2} } ,
$$
where $\partial_a^n=\frac{\mathrm{d}^n}{\mathrm{d}a^n}.$
Remark 2. Another integral could be done by series manipulation
$$
\begin{aligned}
&\int_{0}^{1}K^\prime(k)\left ( \frac{1}{\sqrt{1+ak} }
+\frac{1}{\sqrt{1-ak} } \right ) \text{d}k\\
=& 2\sum_{n=0}^{\infty} a^{2n}\frac{\left ( \frac12 \right )_{2n} }{
(1)_{2n}} \int_{0}^{1}x^{2n} K^\prime(x)\text{d}x=\frac{\pi^2}{2}\,_3F_2\left ( \frac14,\frac12,\frac34;1,1;a^2 \right ).
\end{aligned}
$$
Let $a=1/3$,
$$
\int_{0}^{1}K^\prime(k)\left ( \frac{1}{\sqrt{3+k} }
+\frac{1}{\sqrt{3-k} } \right ) \text{d}k=\frac{1}{48\pi\sqrt{2} } \Gamma\left ( \frac1{24} \right )
\Gamma\left ( \frac5{24} \right ) \Gamma\left ( \frac7{24} \right )
\Gamma\left ( \frac{11}{24} \right ).
$$
That's why I believe an alternative proof exists.
