Yes, when they say "very long" they mean infinite. So the field is purely radial, there is no axial component. In this case the flux through the plane faces (the "lids" of the cylinder) is zero because the field is parallel to the surface. There is only non-zero flux through the curved (cylindrical) surface.

For a finite cylinder or cylindrical shell the field does have axial components and the flux through the lids of the gaussian cylinder is not zero. And is not easy to calculate so using Gauss' law to find the field isnot useful for this situation.

If the Gaussian cylinder is of a finite length L and encompasses a subsection of the long cylinder of charge, there should be an electric field radiating perpendicular to the caps. That calculation is not included to the solution of the problem I sent you.

This seems to go nowhere. I repeatedly explained that if the cylinder of charge is infinite the field is purely along the radius (radial), which means perpendicular to the axis of the cylinders. Both cylinders. So perpendicular to the axis means parallel to the caps. So no flux throuh the caps. I don't knw what more to say. The gaussian is always a finite cylinder but the geometry of the field depends on the cylinder of charge being (or not) infinite.

Yes, I understand what you just said. You're not understanding me. I am saying that if the Gaussian cylinder contains the infinite cylinder, there should be an electric field perpendicular to the caps since the Gaussian surface has finite lenght.

The Gaussian is finite. It has a length L, finite. It cannot contain the infinite cylinder. Just look at any image accompanying such a case.

I mean contain a subsection of the infinite cylinder

I don't know what do you mean. The field is produced by the whole (infinite) cylinder and it has radial symmetry. The gaussian is an imaginary surface which does not change the field in any way.

It is not the field of only the subsection you are calculating but the field of the whole infinite charged wire or cylinder. This so for any use of a gaussian surface. The fact that the flux through the gaussian is equal to the charge enclosed does not mean that the field is produced only by the charge enclosed.

You can see this ifyou take an arbitrary gaussian that does not include any charge. The flux through the surface will be zero but the field will not be so, if there are any charges outside the gaussian surfaces.