I like to think of helices on cylinders as images of lines under planar curling, so here's an approach incorporating that idea.
Imagine the circle rolling up a line drawn in the plane, with $P$ the point of tangency along the line and $Q$ a distinguished point on the circumference of the circle. (So, $Q$ traces out a cycloid in the plane.) Now, imagine the plane is made of thin paper, but the circle is made of stiff cardboard. If we curl the plane into a cylinder but the circle remains flat, the plane of the circle will be tangent to the cylinder along the (vertical) line passing through the point $P$. The point $P$ will follow a helix, and the point $Q$ --which lies in the plane of the circle-- traces out the helical cycloid in space. While the path that $P$ (and $Q$) takes is decidedly different after curling than before, the displacement vector between $P$ and $Q$ in the plane of the circle is the same throughout.
Before curling, if the circle of radius $r$ rolls along a horizontal track in the $uv$-plane, then we have $P=(r t, 0)$, and the point $Q$ traces out the standard cycloid:
$$\begin{align}
u &= r ( t - \sin t ) \\
v &= r ( 1 - \cos t )
\end{align}$$
The displacement vector from $P$ to $Q$ at time $t$ is given by
$$d := PQ = [m, n] = r[-\sin t, 1 - \cos t ] = - 2 r \sin\frac{t}{2} \; [\cos\frac{t}{2},-\sin\frac{t}{2}]$$
Rotating the plane through an angle, say, $\theta := \rm{atan\frac{c}{a}}$, and writing $b$ for $\sqrt{a^2+c^2}$, the circle's point of tangency along the tilted track is given by ...
$$P_2 = (r t \cos\theta,r t \sin\theta) = \left( \frac{r t a}{b}, \frac{r t c}{b}\right)$$
... and the displacement vector, $d_2$, to the point $Q_2$ tracing out the tilted cycloid is given by ...
$$\begin{align}
d_2 :&= [m \cos\theta - n \sin \theta, m \sin\theta + n \cos \theta ] \\
&= \frac{1}{b} [a m - c n, c m + a n] \\
&= -\frac{2r}{b}\sin\frac{t}{2}\left[\cos\left(\theta-\frac{t}{2}\right), \sin\left(\theta - \frac{t}{2} \right) \right] \\
\end{align}$$
Now the fun part: Curl the $(u,v)$-plane around a cylinder of radius $a$, such that the $uv$-origin aligns with the $xyz$-point $(a,0,0)$ and the $v$-axis runs parallel to the $z$ axis. The curled, tilted track will coincide with a helix. The horizontal distance travelled by $P_2$ in the plane becomes the length of horizontal circular arc travelled by $P_3$ around the cylinder; upon dividing by the radius, $a$, of the cylinder, this length becomes the angular "distance" --$s := \frac{rt}{b}$-- travelled by $P_3$; the vertical distances match. Therefore:
$$P_3 = \left(a\cos s, a\sin s, \frac{rtc}{b} \right)=(a\cos s,a\sin s,c s)$$
As for the displacement vector: The $u$ direction of the tangent plane coincides with the horizontal vector tangent to the cylinder at $P_3$; the $v$ direction coincides with the $z$ direction. Thus, the transformation from $uv$-coordinates to $xyz$-coordinates is given by
$$[1,0]\to[-\sin s, \cos s, 0]\hspace{0.5in}[0,1]\to[0,0,1]$$
The image, $d_3$, of the displacement vector $d_2$, then, is
$$d_3 = -\frac{2r}{b}\sin\frac{t}{2}\left[-\sin s \cos\left(\theta-\frac{t}{2}\right), \cos s \cos\left(\theta-\frac{t}{2}\right), \sin\left(\theta - \frac{t}{2} \right) \right] $$
and the path of $Q_3$, which traces the helical cycloid, is given by
$$
Q_3 = P_3 + d_3 = \left\{
\begin{align}
x &= a \cos s &+ \frac{2 r}{b} \sin s \sin\frac{bs}{2r} \cos\left(\theta-\frac{bs}{2r}\right)\\
y &= a \sin s &- \frac{2 r}{b} \cos s \sin\frac{bs}{2r} \cos\left(\theta-\frac{bs}{2r}\right)\\
z &= c s &- \frac{2 r}{b} \sin\frac{b s}{2r}\sin\left(\theta-\frac{bs}{2r}\right)
\end{align}
\right.
$$
Here's a picture with $a=c=1$ and $r=1/2$:
Note: The above does not simply curl the planar cycloid around the cylinder. Since
$$x^2 + y^2 = a^2 + \frac{4r^2}{b^2} \sin^2\frac{bs}{2r} \cos^2\left(\theta-\frac{bs}{2r}\right) \ge a^2$$
we see that most of the helical cycloid lies outside the surface of the cylinder.
Best Answer
I think the problem can be solved by using the tube around the helix from an old answer and then placing the two wires on diametrically opposite sides of that tube to prevent contact.
Here's what it looks like. First from the side:
Then a view from above:
On with the explanation. There is a helix curve in between the red and blue wires in my image. It has the usual parametrization $$ \vec{r}=(R\cos t, R \sin t, ht), $$ where $R$ is the radius of the tube the helix sits on, and the curve will rise $2\pi h$ per revolution (using a negative $h$ reverses the handedness of the screw).
Differentiation w.r.t. $t$ gives the tangent vector along the central helix $$ \vec{t}=(-R\sin t, R\cos t, h). $$ We record that this has constant length $\sqrt{R^2+h^2}$. We see that the derivative of the tangent, after normalizing to unit length, gives $$ \vec{n}=(-\cos t,-\sin t,0) $$ as the normal (perpendicular to $\vec{t}$). The binormal (again normalized) is the cross product $$ \vec{b}=\frac1{||\vec{t}||}\vec{t}\times\vec{n}=\frac1{\sqrt{R^2+h^2}}(h \sin t,-h\cos t,R). $$ We see that $\vec{n}$ and $\vec{b}$ are perpendicular to each other and to $\vec{t}$. Therefore the points given by vectors like $$\vec{r}+ a\cos u\vec{n}+a\sin u\vec{b}\qquad(*)$$ $0\le u\le 2\pi$ form a circle centerd at $\vec{r}$ and perpendicular to the helix. By also varying $u$ they thus parametrize that tubular surface. We also see that by replacing $u$ with $u+\pi$ we get two points on the opposite sides of the helix. We see that the same effect is gotten by replacing $a$ with $-a$.
To get the wires intertwined around the central helix you want $u=k t$ for some parameter $k$. This is thus my suggested answer:
wire number 1: $$ \vec{r}+a\cos(kt)\vec{n}+a\sin(kt)\vec{b}, $$ wire number 2: $$ \vec{r}-a\cos(kt)\vec{n}-a\sin(kt)\vec{b}. $$ This way they are separated from each other by a constant distance $2a$ (assuming that $2a$ is smaller than the rise per revolution $2\pi h$).
When $R$ is much bigger than $h$ you may get away with using $\vec{b}'=(0,0,1)$ instead of the 'technically correct' binormal $\vec{b}$ I used. The distortion won't probably be much. It seems to me that the formulas you showed do just that.
In my images, $R=10$, $h=1/\pi$, $a=0.6$, $k=4$.
Below there is a small section of the above image with a partially transparent (
Opacity[0.6]in Mathematica) tube between the red and blue wires.