This proof is correct, and it is the natural way to argue, given the inputs that
you have. In some sense the indices are also natural: they encode all the relevant data.
If you want to remove some of them, though, here is one standard approach:
First assume that $G$ has $p$-power order, and prove the result in this case.
(I.e. prove your Claim first.) This eliminates your index $i$ in this part of the argument. (Note by the way that your indices $\beta$ should actually be decorated with $i$ as well as $j$, but in this approach they don't need to be.)
Now explain how to deduce the general case from the $p$-power order case. (This amounts to joining together more-or-less the first and last paragraphs of your proof. Now you need the index $i$, but you don't need the $\beta$s or $\delta$s, because they were only used in the proof of the claim.)
I call this a "standard appraoch" because
reorganizing steps of a proof so that various claims, etc., get proved first is a standard method for avoiding an overgrowth of notation. Ultimately, this is often why steps of the proofs of theorems are broken up into lemmas.
Let me elaborate on Tsemo Aristide's anwser that the direct sum of abelian groups should be replaced by the so-called free product for non-abelian groups.
When we want to compare a construction from the theory of abelian groups with the situation of general groups, we first need a way of comparing the class of abelian groups with the class of groups in general. The usual way of doing this is to use categories. A category is just a collection of objects, together with special 'maps' between them (note that I'm glossing over a whole lot of details here, but the Wikipedia page on categories has a lot of information). For instance, there is the category of groups (with group homomorphisms as maps), the category of $\mathbb{R}$-vector spaces (with linear maps), the category of sets (with no restriction on the maps), etc.
In our case, we want to compare the category of groups with the category of abelian groups, and somehow transfer the notion of a direct sum from the latter to the former. To do that, we first need to express the direct sum in the language of categories, i.e. just refering to objects (abelian groups) and maps (group homomorphisms), and in particular without refering to elements of the groups involved. That may seem close to impossible at first glance, but it can be done using the following trick.
First, we notice that if $(A_i)_{i\in I}$ is any collection of abeliqn groups, then for any other abelian group $B$ there is a very natural (set-theoretic) bijection
$$\textrm{Hom}(\oplus_{i \in I} A_i, B) \to \prod_{i \in I}\textrm{Hom}(A_i, B),$$
where $\oplus_{i \in I} A_i$ is the direct sum of the $A_i$ (you should at this point take the time to write down which map this is, and convince yourself that this map is indeed bijective). In words, this says that a homomorphism from $\oplus_{i \in I} A_i$ to $B$ and a collection of morphisms from every $A_i$ to $B$ are 'the same thing'. In fact, it turns out that the direct sum is unique with this property: if $A$ is any abelian group such that for any abelian group $B$ we get a natural bijection like above, then $A$ is isomorphic to $\oplus_{i \in I} A_i$. So this gives us precisely a characterisation of the direct product in terms of just the groups and the homomorphisms.
Now that we know what a direct sum looks like in the category of abelian groups, we can try doing the same for general groups. So given a collection $(G_i)_{i \in I}$ of not necessarily abelian groups, we want to find a group $G$ with the property that for any group $H$, there is a natural bijection
$$\textrm{Hom}(G, H) \to \prod_{i \in I}\textrm{Hom}(G_i, H).$$
Based on the abelian case, we might expect that this group $G$ should be the direct product of the $G_i$, or perhaps the subset of the direct product $G_i$ of sequences with only finitely many non-identity components, but unfortunately these groups do not satisfy the above condition (you should verify this by finding some explicit examples of sets $I$ and groups $G_i$ and $H$ for which we do not have a bijection like above). The group that does satisfy the condition is the one called the free product of the $G_i$. This group is formed by considering finite words, where the letters are taken from all the sets $G_i$, and we may simplify words by multiplying two adjecent letters if they come from the same $G_i$, and we may remove any identity elements from the words (see the wiki page for a more precise definition). It is a nice exercise to show that in the case that all the $G_i$ are abelian, the direct sum of the $G_i$ is isomorphic to the abelianization of the free product, so the free product is indeed a generalization of the direct sum.
So we see that the 'correct' translation of the direct sum concept to general groups leads to free products, instead of cartesian products or subsets thereof. In other words, there is not a problem with considering 'direct sums' of non-abelian groups per se, but to preserve the properties that the direct sum has in the abelian case, we need to consider the free product instead of a direct sum.
Best Answer
There are several problems with what you write, although it can be made to work by taking a few steps "back."
First: it is not true that if you pick an arbitrary generating set for a finitely generated abelian group $G$, say $g_1,\ldots,g_n$, then you will necessarily have that $G$ is the direct sum of the cyclic groups generated by the $g_i$; even if you pick your set to be minimal. For example, in $G=\mathbb{Z}$, then $g_1=2$ and $g_2=3$ generate, no proper subset of $\{g_1,g_2\}$ generate, but $G$ is not isomorphic to $\langle g_1\rangle \oplus \langle g_2\rangle$.
It is true that one may select a suitably chosen generating set with that property, but this fact is not immediate or immediately obvious.
Second, even if you know that $G=\langle g_1\rangle\oplus \cdots \oplus \langle g_n\rangle$, it does not follow that if $H$ is a subgroup of $G$ then you can write $H=H_1\oplus \cdots \oplus H_n$ with $H_i$ a subgroup of $\langle g_i\rangle$. For example, the diagonal subgroup $H=\{(n,n)\in\mathbb{Z}\oplus\mathbb{Z}\mid n\in\mathbb{Z}\}$ of $\mathbb{Z}\oplus\mathbb{Z}$ is not equal to the direct sum of a subgroup of $\{(n,0)\mid n\in\mathbb{Z}\}$ and a subgroup of $\{(0,m)\mid m\in\mathbb{Z}\}$.
The following is true, however:
Taking this for granted, let $G$ be a finitely generated abelian group. Let $X$ be a generating set. Then $G$ is a quotient of a free abelian group $F$ of rank $n=|X|$, $G\cong F/N$.
If $H$ is a subgroup of $G$, then $H$ corresponds to a subgroup $K$ of $F$ that contains $N$, with $H\cong K/N$. By the theorem, $K$ is generated by $r\leq n$ elements, and therefore so is $K/N$. So $H$ is finitely generated.
As for examples in the nonabelian case, I'm not sure if your idea with $D_n$ will work; notice that composing reflections can yield a rotation! For instance, in $D_4$, the relection of the square about the $x$ axis composed with the reflection about the $y$ axis results in a rotation, not a reflection. So you aren't just going to get "the reflections", you are going to get the whole of $D_{2n}$.
For an example you can get your hands on, consider the group $G$ of the $2\times 2$ invertible matrices generated by $$ \left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right) \qquad \text{and}\qquad \left(\begin{array}{cc}2 & 0\\ 0&1\end{array}\right),$$ and let $H$ be the subgroup of elements of $G$ whose main diagonal entries are both equal to $1$. Verify that $H$ is a subgroup of $G$ that is \textit{not} finitely generated.