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 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 −log10 of the acid dissociation constant Ka) 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.
CH4 on the other hand has a pKa of about 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 CH4 is too high, relative to water, meaning that CH4 is a much weaker acid.
The equilibrium lies on the side of the weaker acid, so CH4 will want to remain as CH4.
But, this largely depends on the solvent. If these are placed into water, then yes, HF dissociates partially, giving an acidic solution, and CH4 won't do anything at all.
