No, a "sum of subgroups $H_i$" is the subgroup generated by the subgroups, which consists only of all finite sums of elements of $\cup H_i$; there is no such thing as "infinite sums" in general groups (you need some notion of convergence in order to talk about infinite sums).
In particular, the abelian group $\mathbb{Q}$ does not contain infinite sums. With repeated application of a binary operation you can get sums with arbitrarily large but finite number of summands, not infinite sums/series.
There are two types of "direct sum". The so-called "External Direct Sum" (which is a subgroup of the direct product, etc), and the so-called "Internal Direct Sum".
An abelian group $G$ is the external direct sum of groups $\{G_i\}_{i\in I}$ if $G$ is exactly the set of $I$-tuples, $\prod_{i\in I}G_i$, with almost all entries trivial.
If, on the other hand, $\{H_i\}_{i\in I}$ are a family of subgroups of $G$ such that $\sum H_i = G$ and $H_i\cap\left(\sum_{j\neq i}H_j\right)=\{0\}$, then $G$ is isomorphic to $\oplus H_i$, but is not actually equal to this direct sum. Then we say that $G$ is the internal direct sum of the $H_i$.
Technically, internal and external direct sums are different types of constructions. Morally and in practice, they are the same, because if $G$ is the internal direct sum of its subgroups $H_i$, then it is isomorphic to the external direct sum of the $H_i$; and the external direct sum of the $H_i$ is an internal direct sum of the subgroups $\mathcal{H}_i$, where $\mathcal{H}_i$ is the subgroup of all elements whose $j$th entries are equal to $0$ for all $j\neq i$. So in practice, the distinction is dropped.
Robinson is perfectly correct in saying that $T$ is the sum of the subgroups (i.e., the subgroup generated by them). You are, also, correct in noting that this sum is in fact "direct" (that is, that $T$ is the internal direct sum of the $p$-torsion parts of $G$). If a group is the internal direct sum of some subgroups, then it is also the sum of these subgroups (the converse does not hold, if there are intersections between the subgroups).
Take the infinite dihedral group $D_{\infty} = \{r,s\mid s^2=1, sr=r^{-1}s\}$. The elements of $D_{\infty}$ are precisely the elements of the form $r^is^j$, with $i$ arbitrary and $j=0$ or $1$. The elements of finite order are precisely the identity, and those of the form $r^is$, which have order $2$. Thus, $T=\{r^is\mid i\in\mathbb{Z}\} \cup \{e\}$.
Since $r^i = s(r^{-i}s)\in sT$, then we have that $D_{\infty}=T\cup sT$. But $D_{\infty}$ is not a torsion group, since $r\in D_{\infty}$ has infinite order.
$D_{\infty}$ can also be realized as the semidirect product $\mathbb{Z}\rtimes\mathbb{Z}_2$, where the action is by inversion.
Note that this is an easy example that shows that the set of torsion elements need not be a subgroup in a nonabelian group (though it generates a characteristic subgroup since as a set it is invariant under automorphisms).
Best Answer
For your example, $T$ is bounded, so to apply 4.3.8, it suffices to show $T$ is pure.
Let $t\in T$, and suppose $n{\,\mid\,}t$ in $G$.
We want to show $n{\,\mid\,}t$ in $T$.
Since $n{\,\mid\,}t$ in $G$, we have $t=ng$ for some $g\in G$.
Since $t\in T$, we have $mt=0$ for some positive integer $m$, so $$(mn)g=m(ng)=mt=0$$ hence $g\in T$
Thus $n{\,\mid\,}t$ in $T$, so $T$ is pure.
Now consider the general case . . .
Thus suppose $T=D\,{\large{\oplus}}B$, where $D$ is divisible and $B$ is bounded.
Our goal is to show that $T$ is a direct summand of $G$.
Since $D$ is divisible, it follows by 4.1.3 that $G=D\,{\large{\oplus}}H$ for some $H\le G$.
Let $S$ be the torsion subgroup of $H$.
Since $B$ is bounded, we have $mB=0$ for some positive integer $m$.
Let $s\in S$.
Clearly $s\in T$ so $s=d+b$, where $d\in D$ and $b\in B$.$\;$Then \begin{align*} & s=d+b\\[4pt] \implies\;& ms=md+mb\\[4pt] \implies\;& ms=md\\[4pt] \implies\;& ms\in D\\[4pt] \implies\;& ms=0\;\;\;\text{[since $ms\in S\subseteq H$ and $D\cap H=0$]}\\[4pt] \end{align*} Thus $mS=0$ so $S$ is bounded.
Since $S$ is the torsion subgroup of $H$, it follows (as in the first part of this post) that $S$ is pure in $H$, hence since $S$ is bounded, it follows by 4.3.8 that $S$ is a direct summand of $H$.
Thus we have $H=S\,{\large{\oplus}}K$ for some $K\le H$, hence we can rewrite $G=D\,{\large{\oplus}}H$ as $G=D\,{\large{\oplus}}S\,{\large{\oplus}}K$.
Claim $T=D\,{\large{\oplus}}S$.
The inclusion $D\,{\large{\oplus}}S\subseteq T$ is clear.
For the reverse inclusion, let $t\in T$.
Since $G=D\,{\large{\oplus}}S\,{\large{\oplus}}K$, we have $t=d+s+k$, where $d\in D$, $s\in S$, and $k\in K$.
Then from $k=t-d-s$, it follows that $k$ has finite order, hence $k=0$.
Thus $t=d+s\in D\,{\large{\oplus}}S$, which yields $T\subseteq D\,{\large{\oplus}}S$.
Hence $T=D\,{\large{\oplus}}S$, as claimed.
Then we can rewrite $G=D\,{\large{\oplus}}S\,{\large{\oplus}}K$ as $S=T\,{\large{\oplus}}K$, so $T$ is a direct summand of $G$ as was to be shown.