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.
