Answer: because that's $ACA_0$, alright, but:
Friedman et al.'s 1983 "Countable algebra and set existence axioms" has [verbatim, including old terminology and dubious notation]:
Lemma 1.6 ($RCA_0$) (Bounded $\Sigma^0_1$-separation) If $\varphi$ is $\Sigma^0_1$, then for each $n$ there is an $X$ such that
$\forall x < n(x \in X \leftrightarrow \varphi(x))$
and a few pages later
We now pause to introduce within $RCA_0$ two construction principles which will be useful in several places including the proof of the next Theorem.
(1) Suppose $F_0, F_1,...$ is a sequence of countable fields and $\Pi_n: F_n \rightarrow F_{n+1}$ is a sequence of monomorphisms. Then the direct limit (i.e. the union) of this system exists; $lim_{\rightarrow}F_n$ is the set of all pairs $(n, x)$ where $x \in F_n$ modulo the equivalence
$(n, x) \equiv (m, y) \leftrightarrow n<m$ (or vice versa) $\land \ y = \Pi^{0...0}_{m-1}\Pi^0_{n+1}\Pi_n(x)$
and this appears to make it possible more generally to use bounded $\Sigma^0_1$-comprehension to form an 'increasing' sequence of sets $X_1,...,X_n$ and injections so as to form an arbitrary $\Sigma^0_1$ set in $X = lim X_n$; something about this is wrong, of course, but I'm not sure what
This is because the increasing sequence of sets may not exist, that is, $X_1,X_2,\dots$ exists but the sequence as a whole is not a set. Informally we may think that the sets in a model of $RCA_0$ are "computable" (this is almost never the case) and so there is some program that computes $X_i$ for each $i\in\mathbb{N}$. But it might not be possible to have one program that computes all of the $X_i$, that is a program that accepts input $(n,i)$ if $n\in X_i$ and rejects input $(n,i)$ if $n\notin X_i$. If the sequence $(X_i)_{i\in\mathbb{N}}$ exists, in your case, then $\bigcup_{i\in\mathbb{N} }X_i=X$ also will exist. This is because I can verify if $n\in X$ by checking if $n\in X_n$. This is even more clear when you restrict yourself to standard models of second order arithmetic (in the literature these are called $\omega$-models). Any finite set $F=\{x_0,\dots,x_{n-1}\}$ in a standard model is computable by a program that at input $y$ checks if $y=x_0$ or $y=x_1$,$\dots$, or $y=x_{n-1}$. Any set $Z$ of the natural numbers is the union of finite sets, however you cannot piece together the programs of these finite sets to obtain a program that computes $Z$.
It must be pointed out that for arbitrary sequences of sets $\textbf{RCA}_0$ is unable to prove that the union of the sets exists, in fact being able to take arbitrary countable unions is also equivalent to arithmetic comprehension. In fact you have that any $\Sigma^0_1$ definable set $Y$ is the union of a increasing sequence of finite sets $(A_i)_{i\in\mathbb{N}}$. However to check if $n\in \bigcup_{i\in\mathbb{N}} A_n=Y$ i would need to search for an $i\in \mathbb{N}$ such that $n\in A_i$ which is an unbounded search which may not terminate.
Friedman is supposing that the collection of countable fields $(F_i)_{i\in\mathbb{N}}$ is coded by a single second order object, or set of naturals. Informally this means there is a single program that computes all the fields $F_i$ and their respective operations. He also avoids issue with unions since he quotients the sequence by some computable equivalence relation.