One observation is, that the coordinate functions can't be smooth in the corners of the convex hull, so for polynomials as coordinate functions, the only points on the curve that qualify as corners of the CH would be the end points of the curve segments.
That observation yields the sufficient condition, that the inner points of the curve segments must have non-negative barycentric coordinates with respect to the simplex defined by the curve segment's endpoints.
The Oloid and also the Sphericon however are the convex hull of two 3D curves, that are the union of an uncountable number of line-segments without being a polyhedron.
Edit
I will try to formulate some reasonable conditions for two suitable curves:
both curves must be planar, but not co-planar; this condition may however only be a sufficient condition and only cover a special case.
both curves must have finite length and contain the end-points
the containing plane of each curve must intersect the other in an inner point
if the curves are rotated into coplanar position around the intersection of the containing planes, the convex hull of the curves in coplanar orientation consists of the entire two curves and a maximum of two line-segments joining endpoints of the curves.
Edit II
rethinking the question, whether the curves have must be planar and, the generating curves of the Oloid and of the Sphericon in mind, it occured to me that space-curves need not be described parametrically or implicitly; instead they could also be described as the intersection of two appropriately chosen surfaces.
From the Oloid and the Sphericon I concluded that each of the two curves must be the intersection of a pair of generalized cones, with the following
Definition Generalized Cone
let $\gamma(t) = (\gamma_x(t),\gamma_y(t)), t\in[0,2\pi)$ be the boundary of a convex region, that has non-zero, finite area and, that contains the origin.
A generalized cone is then the surface $C_{\gamma}(u,v) = (\gamma_x(u)*v,\gamma_y(u)*v,v), u\in[0,2\pi), v\in[0,+\infty)$
now a pair of the desired curves can be generated as follows:
w.l.o.g. assume that the endpoints of curve $B$ are $B(0)$ and $B(1)$" which can be assumed to coincide with $(0,0,-1)$ resp. $(0,0,1)$
choose two, not necessarily indentical, convex curves $\gamma_{B(0)}(t)$ and $\gamma_{B(1)}$ to be used in the definition of two surface, in whose intersection curve $A$ shall be deemed to be contained. The two surfaces are then $F_{B(0)}(u,v) := (\gamma_{B(0)x}(u)*v,\gamma_{B(0)y}(u)*v,v+1)$ and $F_{B(1)}(u,w) := (\gamma_{B(1)x}(u)*w,\gamma_{B(1)y}(u)*w,-(w+1))$
choose a connected portion of $F_{B(0)}(u,v)\cap F_{B(1)}(u,w)$ and take its endpoints as the tips of the second pair of generalized cones, which after an appropriate invertible linear transformation can also be assumed to correspond to $(0,0,-1)$, resp. to $(0,0,1)$
choose as the convex curves $\gamma_{A(0)}(t)$ and $\gamma_{A(1)}(t)$ the intersection of the "partial generalized double-cone" defined via the transformed curve $A$ and the transformed endpoints $B(0)$ and $B(1)$ of curve $B$ (which is not yet determined) with the xy-plane and augment that planar intersection with curve-segments whose union with the aforementioned intersection with the double-cone constitutes to appropriate convex curves $\gamma_{A(0)}(t)$ and $\gamma_{A(1)}(t)$
that admittedly complicated description makes clear, that the curves need not be planar: take for the generalized double cones as the generating curves non-circular ellipses $x(t) =a*\cos(t), y(t)=b*\sin(t)$ and $x(t)=b*\cos(t), y(t)=a*\sin(t)$
Checking a concrete pair of curves
the set $\gamma(t)$ of points defined by the intersection of the lines $A(s_0)+\lambda*(B(t)-A(s_0)), s_0\in\{0,1\}, 0\le t\le 1$ with a plane that is orthogonal to the line $A(0)+\mu*(A(1)-A(0))$ must be convex in the sense that no line intersects it more than twice.
The same condition must hold after the roles of $A$ and $B$ have been exchanged.$\gamma(t)$ can be augmented to the boundary of a planar convex pointset that contains the intersection of the line through the endpoints of $A$ with the plane containing $\gamma(t)$
The same condition must hold after the roles of $A$ and $B$ have been exchanged.neither of the curves must have a point between the two planes $E_A$ and $E_B$, for which $dist(E_A,A(0))=dist(E_A,A(1))=0, dist(E_A,B(0))=dist(E_A,B(1))=c$, resp. $dist(E_B,B(0))=dist(E_B,B(1))=0, dist(E_B,A(0))=dist(E_B,A(1))=c$.
what may still be missing is a condition that guarantees that the entire convex hull of $A\cap B$ is covered.