Show that $\mathbb{Q}^+/\mathbb{Z}^+$ cannot be decomposed into the direct sum of cyclic groups.

Some comments:

1. You are conflating elements of $\mathbb{Q}$ with elements of $\mathbb{Q}/\mathbb{Z}$. The latter are congruence classes modulo $\mathbb{Z}$. Better to keep them straight.

2. You assert that if $\mathbb{Q}=\oplus H_k$, and $\frac{a}{b}+\mathbb{Z}$ is a generator for one of the direct summands, then there is a direct summand that contains $\frac{1}{b}+\mathbb{Z}$; you have no warrant for that assertion; why can't $\frac{1}{b}+\mathbb{Z}$ have more than one nontrivial coordinate in the direct sum? Better: show that $\frac{1}{b}+\mathbb{Z}\in\langle \frac{a}{b}+\mathbb{Z}\rangle$, by using the fact that $\gcd(a,b)=1$. That will prove that you can always select a generator of the form $\frac{1}{b}+\mathbb{Z}$, which seems to be what you are trying to do in the first place.

3. Your argument is murky and way too complex. It is simpler to show that any two nontrivial subgroups of $\mathbb{Q}/\mathbb{Z}$ must intersect, and so any direct sum decomposition $A\oplus B$ of $\mathbb{Q}/\mathbb{Z}$ must have $A$ or $B$ trivial. Then use your argument to show that the group $\mathbb{Q}/\mathbb{Z}$ is not cyclic in order to finish the proof. Actually, this suggestion works for $\mathbb{Q}$ but no necesarily for $\mathbb{Q}/\mathbb{Z}$; for instance, the Prufer $p$-groups are subgroups of $\mathbb{Q}/\mathbb{Z}$ but they intersect trivially for distinct primes. Sorry about that.

   Instead, use a similar argument to what you had before: if $\mathbb{Q}/\mathbb{Z} = \bigoplus H_k$, select $b_k\in\mathbb{Z}$, greater than $1$ without loss of generality (since $b_k=1$ means the subgroup generated by $\frac{1}{b_k}+\mathbb{Z}$ is trivial and can be omitted), such that $\frac{1}{b_k}+\mathbb{Z}$ generates $H_k$. Then select a particular direct summand, say $H_1$; consider the element $\frac{1}{b_1^2}+\mathbb{Z}$ of $\mathbb{Q}/\mathbb{Z}$. Since it is an element of $\mathbb{Q}/\mathbb{Z}=\oplus H_k$, then we can express it as: $$\frac{1}{b_1^2}+\mathbb{Z} = \sum_k\left(\frac{a_k}{b_k}+\mathbb{Z}\right)$$ for some $a_k\in\mathbb{Z}$, with $b_k|a_k$ for almost all $k$ (since almost all components must be trivial). Then adding it to itself $b_1$ times we get that $$\frac{1}{b_1}+\mathbb{Z}=b_1\left(\frac{1}{b_1^2}+\mathbb{Z}\right) = b_1\sum_k\left(\frac{a_k}{b_k}+\mathbb{Z}\right) = \sum_k\left(\frac{a_kb_1}{b_k}+\mathbb{Z}\right).$$ Therefore, equating components, we get that $b_k|a_kb_1$ for all $k\neq 1$, and $b_1|a_1b_1-1$. But the latter implies $b_1|1$, which is a bit of a problem.

1. The group $\mathbb{Q}$ is divisible.

2. A homomorphic image of a divisible group is divisible.

3. The group $\mathbb{Q}/\mathbb{Z}$ is divisible (by 1 and 2).

4. A direct summand of a divisible group is divisible (by 2).

5. No nonzero cyclic group is divisible, because nonzero divisible groups are infinite and $\mathbb{Z}$ is not divisible.

Therefore no nonzero cyclic group is a direct summand of $\mathbb{Q}/\mathbb{Z}$.