Can Reduction for Undecidability/Decidability Problems Go Both Ways?
When proving decidability of a new problem, you reduce the new problem
(i.e. a problem with unknown solution) to an old one (used as
subroutine) with a known solution, so that the techniques used to
solve the old one can be used to solve the new one.

But, when proving undecidability, you have to do the opposite. You
reduce the old problem, known to be undecidable, to the new problem
(used as subroutine), in such a way that, if the new problem has a
decision procedure, its use as a subroutine provides a decision procedure to the old
problem, known not to have any. Hence the new problem cannot have a
solution, and is undecidable.

In your example, you should have tried to show that if you have a
decision for your new problem, that you can use as subroutine, then
you ave a decision for the halting problem, which is not possible.

So proofs go both ways only in the sense that the contrapositive of a statement goes the other way. Decidability goes one way, and undecidability goes the other way.

However, it you can do the reduction both ways at the same time (either both for decidability or both for undecidability), then you have proved that the two problems are strictly equivalent, and that any solution for one is a solution for the other. When it goes only one way, this establishes that one problem is harder than the other (letting you work out which is which).

I guess that is about all I can say about the "direction"of the proof.