I have to admit I have no intuitive feeling (beyond the definition) as what means geometrically to not have a 4-handle for a 4-manifold. Why is it that extending to the 3-skeleton is the same as extending to M ? Is it because a 4-manifold without 4 handle is equal to its 3 skeleton ?

Thank you for you answer, this answers positively 1 and 2, in the case of 4-manifolds without 3-handles. I was not able to directly see this result in the references you mentioned (I guess it's a direct consequence, but I'm not familiar enough with this subject to see that as trivial). So just so that I understand a little bit better, why having no 3-handles implies that there is a canonical bijection between Spin^c structures and almost-complex structures?

Thank you, so I guess the answer to 1. is NO. But it is YES, up to a nullhomotopy choice. I re-worded question 3.

For 3. I can't see how this answers the question, if M has no spin structure, then the answer to "Does every spin structure on M, etc." is yes.

Thank you for your answer. For 1., the article ncatlab.org/nlab/show/spin%5Ec+structure refers to a "canonical" spin^c structure, but I can't say I understand really. Do you see in there a way to choose a special class of nullhomotopy of the composite map ?

Thank you for you anwers. I don't understand the last sentence of your first comment : "this means if $M$ is parallelizable".