Is the hybridization only related to atomic orbitals but not to the molecular orbitals?
Hybridisation is a concept devised by man to ease explanation of chemical observations. It is not a 1:1 equivalent of anything seen in molecules but rather a way of rationalising what one has seen — one that is as mathematically correct as molecular orbitals (that span the entire molecule) are.

There are two fundamentally different approaches on how to describe ‘bonds’. The first one is the molecular orbital approach. It considers electrons that can freely move around the entire molecule, in a so-called potential energy space: the potential energy derives from the (negatively charged) electron being close to (a series of positively charged) nuclei which attract each other. As you may remember from your very first chemistry classes ever, the mass of even a proton is $2000$ times that of an electron, so a carbon nucleus is some $24000$ times heavier than an electron. According to Newton, the nuclei have a lot more inertia than the electrons do, so it is reasonable to assume that whichever fixed positions the nuclei take, all electrons can then assume favourable movement patterns.

Quantum mechanics, and further quantum chemistry, give us methods to calculate what movement patterns these electrons might assume — although technically the equation is not solveable, because it also must contain electron-electron interactions, there are good approximations out there. The wave functions that one can derive from all this information correspond to molecular orbitals you might have seen before: They abide by the molecule’s symmetry (will always fit into one irreduceable representation of the molecule’s symmetry point group) but extend over the entire molecule (even the orbitals representing atomic core electrons can be thought of being delocalised somewhat).

Using the quantum mechanical method of calculation (operators) you get an energy for each molecular orbital. Thereby you have ordered your molecular orbitals by energy (approximately equalling the number of nodal planes) — this gives you HOMO and LUMO that you might have heard of before.

To reduce this back into ‘bonds’, we need to use mathematical tricks. Simply spoken: All wave functions can be treated as vectors, and vectors can be linear combinated to give other vectors — meaning we take $0.5$ times orbital $\Psi_1$, $0.01$ times orbital $\Psi_2$ and so on for each molecular orbital, add them up, divide them by a normalisation factor, and we get a new orbital that has a different shape (and a new energy). If we do this in a sensible way, we can manage to reduce molecular orbitals to those that only ‘span’ across one bond — because we have cancelled out all wave functional contributions from other parts of the molecule. This would correspond to your typical $\sigma$ and $\pi$ type bonding, nonbonding and anti-bonding orbitals.



The second method is the hybridisation approach. This time, you take the general shape of the molecule and consider each atom by itself. For each atom, you can assume a hybridisation based on its geometry: If there are three other atoms in a triangular shape around it, all sharing the same plane with the atom in question, we can assume it to be sp² hybridised.

Once we have done this, we need to take a look at neighbouring atoms to create ‘bonds’ again. Usually, there will be two hybrid orbitals ‘pointing’ at each other, ready to create a bond. Using the same technique we have to create the hybrid orbitals, we can mix these two hybrid orbitals to create another orbital. This orbital is not considered to be of a new sp-hybrid type, but rather gets a new type of name: It can be $\sigma$, $\pi$ or $\delta$ (to name the most common ones). These Greek letters closely resemble s, p and d, and the bond orbitals in question share a common symmetric trait with the corresponding atomic orbitals: $\sigma$ orbitals can be rotated freely around the bond axis, $\pi$ orbitals have one and $\delta$ orbitals two nodal planes along the bond axis (the bond axis is completely part of the nodal plane).

If we did this correctly, we will arrive at orbitals of similar shape and energy as with the above method, only we are much, much faster, which is why this method is often taught in schools.

Onto the question of how do the $\ce{C-C}$ (and $\ce{C-H}$) bonds differ in $\ce{C2H4}$ and $\ce{C2H2}$:
You just saw how we decided what orbitals we want to combine to form a bond. In $\ce{C2H4}$, we are combining a carbon’s sp² orbital with a hydrogen’s s-orbital to create a $\sigma$ bond. For simplicity, we can label this bond $\sigma_{\ce{sp^2-s}}$. Similarly for $\ce{C2H2}$, only that this one has to be labelled $\sigma_{\ce{sp-s}}$. You see that you created bonds with different hybrid orbitals and therefore the bonds will have different ‘shapes’ and energies. This is what is meant with


  The $\ce{C#C}$ bond of ethyne ($\ce{HCCH}$) has higher s character ($50\,\%$) than the $\ce{C=C}$ bond of ethylene ($\ce{H2CCH2}$) ($33\,\%$)


Technically, this is short for:


  The hybrid atomic orbitals of the carbons of ethyne have a higher s character ($50\,\%$) than those of ethene ($33\,\%$), so the $\sigma_{\ce{C-H}}$ bonds will retain that property somewhat.




Long story short: No, we cannot say that a $\sigma$ bond has sp² or sp³ character; we can (and should) only say that the atomic orbital of one side of that bond has this character.

And: It is wrong to assume $\ce{sp^2 + s} = \ce{2 sp}$, because that would recentre a p-orbital from the carbon to the hydrogen.