Why intersection of chords form a cardioid? 

Image from Wikipedia

Put equally spaced points in a circle and label them 1,2,3,4,.. and so on.
Connect 1 to 2, 2 to 4, 3 to 6 and generally $n$ to $2n$.
The intersection of these chords will form a cardioid as shown in the above picture.
Cardioids can also be made from rolling a circle over other and tracing this point.


Image from Wikipedia

Why doing these seemingly different things give the same result?
How is these two operations related?

Mathologer's YT video on Cardioids and multiplication table
Wikipedia article on Cardioid

 A: First, a bit of animated inspiration:

Now, we consider the origin-centered unit circle $\bigcirc O$, about which rolls unit circle $\bigcirc O'$. For $\theta$ the counter-clockwise direction angle of $\overline{OO'}$, we can define $P_\theta$ as the varying tangent point of the circles; $Q_\theta$ as the proverbial "spot of paint" point on $\bigcirc O'$ that traces the cardioid; and $R_\theta$ as the point of $\bigcirc O'$ from $O$, which traces a circle of radius $3$ about $O$. (Note that $P_0=Q_0=(1,0)$.) Let $T_\theta$ be the point where the extended segment $\overline{R_\theta Q_\theta}$ again meets the big circle, and let $S=(-3,0)$ be the fixed point on the far left of the big circle.

By the rolling definition of the cardioid locus, we know that $\angle P_\theta O P_0=\angle Q_\theta O' O$; equivalently, $\angle R_\theta OS = \angle Q_\theta O'R_\theta$ (as marked in the figure). Light angle-chasing through isosceles triangles tells us that $\angle T_\theta O S = 2\angle R_\theta O S$; thus, thinking clockwise from $S$, we see that $\overline{R_\theta T_\theta}$ is the precisely type of chord described in the definition of the cardioid as the envelope of chords.
"All we have to do", then, is show that chord $\overline{R_\theta, T_\theta}$ is tangent to the locus of $Q_\theta$. This takes a little Calculus.
The parametric form of our $Q_\theta$ (which, note, is shifted one unit right of the standard cardioid locus) is
$$Q_\theta = \left(\;1 + 2(1-\cos\theta)\cos\theta\;,\;2(1-\cos\theta)\sin\theta\;\right)$$
so the tangent vector has the form
$$\overrightarrow{Q'_\theta} \;=\; \frac{d}{d\theta}Q_\theta \;=\; 2\;\left(\;-(1 - 2 \cos\theta) \sin\theta\;,\;(1 - \cos\theta) (1 + 2\cos\theta)\;\right)$$
Observe that, since $P_\theta=(\cos\theta,\sin\theta)$, we can calculate
$$\overrightarrow{P_\theta Q_\theta}\;=\;Q_\theta-P_\theta \;=\;\left(\;(1 - \cos\theta) (1 + 2\cos\theta)\;,\; (1 - 2 \cos\theta) \sin\theta\;\right)$$
which amounts to the "negative reciprocal" of (half of) $\overrightarrow{Q'_\theta}$. The vectors are orthogonal. But $\overline{P_\theta Q_\theta}\perp\overline{R_\theta T_\theta}$ (by Thales), so it must be that $\overrightarrow{Q'_\theta}$ and $\overline{R_\theta T_\theta}$ are parallel; that is, the chord is indeed tangent to the locus, as desired. $\square$
A: The red curve is a caustic, i.e. the envelope of all rays reflected by a "mirror" (in this case the given circle $c$, of centre $O$ and radius $r$) and issued from some fixed point (in this case point $P$, which is on $x$ axis on the left in the first figure of the question, and on the right in figure below). In fact, one may check that a ray issued from $P$ and hitting $c$ at point $n$ will be reflected to point $2n$: just remember that a ray hitting the circle at some point $A$ is reflected to a ray which is its reflection about line $OA$.
I'll find the geometric properties of this envelope in a purely geometric way, without trigonometry nor calculus, and I'll show it is indeed a cardioid.
Take a point $A$ and a point $A'$ near to it, both on $c$. Let $s$ and $s'$ be the rays reflected of $PA$ and $PA'$, and $Q$ their  intersection point. In the limit $A'\to A$, point $Q$ tends to a point $Q_0$: the envelope is the locus of all $Q_0$, as $A$ varies on the circle.
To find $Q_0$, observe first of all that $\angle AQA'=3APA'$ (see figure below). This is a simple matter of angle chasing: if $\alpha=\angle APA'$ and $\theta=\angle POA$, then
$\angle PAQ=2\angle PAO=\pi-\theta$ and $\angle PMA=\theta-\alpha$, while $\angle AOA'=2\alpha$ and
$\angle PA'Q=2\angle PA'O=\pi-\theta-2\alpha$. It follows that
$$
\angle MQA'=\pi-\angle MA'Q-\angle QMA'=
\pi-(\pi-\theta-2\alpha)-(\theta-\alpha)=3\alpha.
$$

Consider now the circle $c'$ through $QAA'$, of radius $r'$, and arc $\gamma'=AA'$ on that circle. If $\gamma$ is the arc delimited by $A$ and $A'$ on circle $c$, then:
$$
{\gamma'\over\gamma}=
{2r'\cdot\angle AQA'\over 2r\cdot\angle APA'}=
3{r'\over r}.
$$
In the limit $A'\to A$ circle $c'$ tends to a circle internally tangent to $c$, intersecting reflected ray $s$ at $Q_0$. In the same limit, we have $\gamma'/\gamma\to1$, because the ratio between an arc and the subtended chord tends to $1$ when the arc tends to zero; the above equality implies then $r'/r\to1/3$. Circle $c'$ tends then to the circle tangent to $c$ at $A$, having radius $r/3$ and centre $O'$. Point $Q_0$ is thus the intersection between this circle and the reflected ray $s$ (see figure below).
When $A=P$, incident and reflected rays lie on the tangent to $c$ at $P$, hence $Q_0=P$. The internally tangent circle $i$ is also tangent to a circle with the same radius, concentric to $c$. If we imagine to roll $i$ on this inner circle, after a rotation of angle $\theta=\angle POO'$ we also have $\angle AO'Q_0=\angle POA=\theta$. Hence $Q_0$ can also be obtained from rolling a fixed point on circle $i$. This shows that its locus is indeed a cardioid.

