= Solution
The printed construction is not well defined: the next tangent must pass through $x_1$, rather than $x_0$. If $x_0\ne x_1$ and $\ell_1\ne\ell_0$, a second line through $x_0$ cannot also pass through $x_1$, because the unique line through both points is $\ell_0$. Here is a concrete counterexample satisfying all the conic hypotheses. In the affine chart take
$$
C_1:x^2+y^2=1,\qquad C_2:x^2/4+y^2/9=1,
$$
with $\ell_0:x=1$, $x_0=(1,3\sqrt3/2)$ and $x_1=(1,-3\sqrt3/2)$. The second tangent through $x_0$ is $-23x+12\sqrt3y=31$. It does not contain $x_1$: its left side there is $-77$. The conics meet transversely at the four complex points with $x^2=32/5$, $y^2=-27/5$. Thus this is a defect in the original PDF, not just in its conversion.
For the corrected construction, let $\iota_\ell$ exchange the two points of $C_2$ on a fixed tangent line, and let $\iota_x$ exchange the two tangents to $C_1$ through a fixed point of $C_2$. Both projections from $E$ are degree-two morphisms to a <smooth plane conic>: for the line projection this follows from intersecting a line with $C_2$, which has no line component. Since $E$ is smooth and the ground field has characteristic zero, their quadratic function-field extensions define regular involutions on the whole curve. At a ramification point “the second” point is the same point, counted with multiplicity. Each switch is an <involution of a degree-two map from a genus one curve>. Hence the corrected step is the everywhere-defined automorphism $F=\iota_x\circ\iota_\ell$.
Choose an origin $O$ on the <genus one curve>. The <Abel-Jacobi map of a genus-one curve> identifies $E$ with $\operatorname{Pic}^0(E)$ by $p\mapsto\mathcal O_E(p-O)$. Fibers of each degree-two projection are linearly equivalent <Weil divisors>, since they are pullbacks of points of $\mathbb P^1$. Thus their group sums are constant: for suitable $a,b\in E$,
$$
\iota_\ell(p)=a-p,\qquad \iota_x(p)=b-p,\qquad F(p)=p+(b-a).
$$
This also holds at the ramification points, where $2p=a$ or $2p=b$. Therefore the corrected step is a <translation on an elliptic curve>. If $F^n(p_0)=p_0$ for one point, then $n(b-a)=0$. It follows that $F^n(p)=p$ for every $p\in E$. \b[The corrected construction has the <Poncelet porism>: one periodic orbit implies all orbits are periodic, with the same least period.] The literal printed construction fails before this conclusion; the proof establishes the intended, explicitly corrected assertion.
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