Well, I don't know which kind of Turing machine these are, as it is just moving its head instead of producing some output and there is no language definition given for acceptance.

However, a problem is called decidable if there is an algorithm to decide the set membership problem. Let us say the language accepted by the given Turing machine M is L has then given an input if we can decide in finite time that input belongs to the language (acceptance i.e. halting on an accepting state) or input do not belong to the language (rejection i.e. halting on a rejecting state). That problem would be called decidable.

Remember halting on rejection is also important here if it is looping forever and not halting on a non-accepting input then it is undecidable. And the language which is decidable is called recursive.

We need to get either a 'YES' or 'NO' answer in our membership problem to be decidable, looping forever would not be good enough for rejection.

As far as the given two machines are concerned, in case of A when machine tries to move its head to the left of the input, it really depends on the input, if it is finite then it would be clear that machine is able to do it(YES) or not been able to do it(NO). If the input is infinite machine would not halt hence undecidable.

When we take the case of B which moves its head to the left at least once, it is clearly decidable, no matter whether the input is finite or infinite, it would either be successful in moving to the left once or not i.e. it will halt.