To ionize is to dissociate, so that aside...
"pKa" is a nice relative measure of acidity. The higher the "pKa", the weaker the acid (and the stronger its bonds with its "H"^(+)).
"HF" is an acid; it has an \mathbf("H"^(+)) on it, since the anion is "F"^(-). Therefore, it can dissociate to donate an "H"^(+) to solution, making it follow (at least) the definition of a Bronsted acid and that of an Arrhenius acid.
Its "pKa" (the -log_10 of the acid dissociation constant K_a) is about 3.17, and if you place it in water, whose "pKa" is higher, "HF" will be the stronger acid.
The lower pKa indicates the stronger acid.
"CH"_4 on the other hand has a "pKa" of about \mathbf(50), so it's very basic (or very poorly acidic). That means its "H" will never in a million years want to dissociate in water as "H"^(+); its "pKa" is much higher than that of water.
Its "C"-"H" bonds are too strong (thermodynamically stable), because the "pKa" of "CH"_4 is too high, relative to water, meaning that "CH"_4 is a much weaker acid.
The equilibrium lies on the side of the weaker acid, so CH_4 will want to remain as CH_4.
But, this largely depends on the solvent. If these are placed into water, then yes, "HF" dissociates partially, giving an acidic solution, and "CH"_4 won't do anything at all.
