Here is a sketch of proof that there is no geodesic with self-intersections as on the second diagram.

Assume a convex surface $\Sigma$ with such geodesic exists; suppose that arcs and angles are labeled by their lengths.

Apply Gauss--Bonnet formula to show that $$2\cdot\alpha<\beta+\gamma$$ and $$2\cdot\beta+2\cdot \gamma<\pi+\alpha.$$ Conclude that $\alpha <\tfrac \pi 3$.

Consider the part of geodesic without arc $a$. Pass to its convex hull; denote its surface by $\Sigma'$.

Note that $\Sigma'$ is divided by the curve into 4 parts, one pentagon and three monogons. Each of these pieces can be developed in the plane, moreover the resulting figure is convex.

Consider the plane figure that corresponds to the pentagon. Its sides are formed by convex curves with marked lengths, the angles of the pentagon are marked as well. It remains to use inequalities on angles to show that there is no pentagon with these properties.

Here is a sketch of proof that there is no geodesic with self-intersections as on the second diagram.

Assume a convex surface $\Sigma$ with such geodesic exists; suppose that arcs and angles are labeled by their lengths.

Apply Gauss--Bonnet formula to show that $$2\cdot\alpha<\beta+\gamma$$ and $$2\cdot\beta+2\cdot \gamma<\pi+\alpha.$$ Conclude that $\alpha <\tfrac \pi 3$.

Consider the part of geodesic without arc $a$. Pass to its convex hull; denote its surface by $\Sigma'$.

Note that $\Sigma'$ is divided by the curve into 4 parts, one pentagon and three monogons. Each of these pieces can be developed in the plane, moreover the resulting figure is convex.

Consider the plane figure that corresponds to the pentagon. Its sides are formed by convex curves with marked lengths, the angles of the pentagon are marked as well. It remains to use inequalities on angles to show that there is no pentagon with these properties.

Postscript. A complete proof is given in my note "Self-crossing geodesics".