I initially only answered the question whether $H\neq S_R$, where I just confirm I negative answer, and said I think $\mathrm{Sym}_{\mathrm{Rec}}(\mathbf{N})$ is known not to be finitely generated. I've added a proof of this fact as an edit, below.
Original answer:
Your group is easier to describe using $\mathbf{Z}$ than $\mathbf{N}$ (using a recursive bijection between the two). Namely, under this isomorphism it corresponds to $H_1=\langle S_\infty(\mathbf{Z}),f_0\rangle$, and by $f_0$ defined by $f_0(n)=n+1$ for $n\neq 0,-1$, $f(0)=0$, $f(-1)=1$ (infinite cycle with fixed point). This group can be defined "implicitly", namely as the group of permutations of $\mathbf{Z}$ that eventually coincide with a translation. It's also more simply described as $\langle S_\infty(\mathbf{Z}),f_1\rangle$, with $f_1(n)=n+1$.
It's quite clear that $H_1$ is not the whole group of recursive permutations of $\mathbf{Z}$. Indeed, your $f$, viewed as permutation of $\mathbf{Z}$ (fixing negative integers), is not in $H_1$.
(Also there's a unique homomorphism $H_1\to\mathbf{Z}$ mapping $f_1$ to $1$, while $\mathrm{Sym}_{\mathrm{Rec}}(\mathbf{Z})$ can be shown t be a perfect group. Indeed it was proved by C. Kent (1962, link at AMS site) that $\mathrm{Sym}_{\mathrm{Rec}}(\mathbf{Z})$ has only 4 normal subgroups (similarly as in the Onofri/Schreier-Ulam theorem): the whole, the trivial subgroup, the finitary subgroup, and the subgroup of index 2 therein).
Edit: It took me a few hours to reconstitute the argument of infinite generation, which was enough for a grumpy user to downvote.
Lemma There is no computable map $f:\mathbf{N}^2\to \mathbf{N}$ such that $n\mapsto f(n,-)=:f_n$ is a surjection of $\mathbf{N}$ onto $S_R$.
Proof: assume so. Let $u(n,m)=u_n(m)$ be the supremum of $f_n$ on $[0,m]$. So $(m,n)\mapsto u(m,n)$ is computable. Let $u$ be a computable increasing function such that $u\gg u_n$ for all $n$ (it exists by an easy diagonal argument, namely $u=\sum \mathbf{1}_{[n,\infty[}u_n)$). Let $q$ be the permutation exchanging $n$ and $2u(n)$ for every odd $n$, and fixing other elements. Then it is computable, and cannot be among the $f_n$.
Corollary $S_R$ is not finitely generated.
Proof: otherwise, it's generated by some finite subset $S$. Then using the surjective map $p:F_S\to S_R$ and using a computable bijection $q:\mathbf{N}\to F_S$ we get the map $(n,m)\mapsto q(n)m$ which is computable and contradicts the lemma.
$G_{\alpha}$ is described in the first sentence as torsion free abelian, and that persists for the second sentence. In the second part you are considering a finitely generated subgroup $H$ of the direct limit, and you know that the direct limit is torsion free by part one. Subgroups of torsion free abelian groups must be torsion free, so the subgroup is finitely generated torsion free abelian, and therefore (for example by the link I posted earlier) must be free abelian. Choose generators $\{a_1, ..., a_n\}$ for the subgroup $H$. Now you could imitate the proof in the post that you cited: each generator $a_i$ must lie in the image (in the direct limit) of some $G_\alpha$, and since there are finitely many of them, they must all lie in the image of a common $G_\alpha$, and therefore the subgroup itself is in the image of that $G_\alpha$.
Best Answer
Notice that $H$ is an Abelian group, and all its non-identity elements have infinite order. If it were finitely generated, it would be a direct sum of a finite number of copies of $\mathbb{Z},$ say generated by $\{ \left(\begin{array}{clcr} 1 & x_{i}\\0 & 1 \end{array}\right) : 1 \leq i \leq n \}.$ then every element of $H$ could be expressed (uniquely, though that is not essential here) in the form $\left(\begin{array}{clcr} 1 & u \\0 & 1 \end{array}\right)$, where $u = \sum_{i=1}^{n} z_{i}x_{i},$ with each $z_{i} \in \mathbb{Z}.$ Now the $(1,2)$-entry of any $h \in H$ is certainly a rational number ( in fact with denominator a power of $2$). But the denominator of every $\mathbb{Z}$-combination of the $x_{i}$ is at worst the lcm of the denominators of the $x_{i},$ so is bounded. However, $\left(\begin{array}{clcr} 1 & 2^{-i}\\0 & 1 \end{array}\right)$ lies in $H$ for any $i \in \mathbb{N},$ a contradiction.