In Reverse Mathematics, Stillwell states the following:
Realizing a $\Sigma_0^0$ condition by a function. For any $\Sigma_0^0$ condition $\exists m\varphi(m, n)$, there is a function $g:\mathbb{N} \rightarrow \mathbb{N}$ such that $\exists m [g(m) = n] \iff \exists m \varphi(m, n)$.
Proof. ...
Remember that when RCA$_0$ proves existence of a function $g$ this means that the set of ordered pairs $\left\langle n, g(n)\right\rangle$ is computable. We have not proved in RCA$_0$ that the set $\left\lbrace n:\exists m \varphi(m,n)\right\rbrace$, the range of $g$, exists. Indeed, we cannot do this in RCA$_0$ when $\left\lbrace n: \exists m \varphi(m,n)\right\rbrace$ is computably enumerable but not computable.
To claim the existence of the set $\left\lbrace n: \exists m \varphi(m,n)\right\rbrace$ we would need more than recursive comprehension; we would need $\Sigma_1^0$ comprehension....
But I don't understand why this should be the case. Doesn't the generation of pairs $\left\langle n, g(n) \right\rangle$ itself constitute a constructive proof of the existence of $\left\lbrace n : \exists m \varphi(m,n)\right\rbrace$?
-- Or, like, what precisely does it mean for a constructed set to "exist", that we would need to prove it? In cases where we say "Consider a set with the following properties", then the set exists iff our properties are not self-contradicting -- but rather than starting from a high-level property and trying to find a set that satisfies it, at this point after the proof we have some (possibly infinite) list of members and we're just wrapping a pair of curly braces around them and calling it a set, aren't we?
This is a common confusion faced when learning reverse mathematics. There are three different objects to consider here:
The function $g$.
The graph of $g$, $\{\langle n,g(n)\rangle: n\in\mathbb{N}\}$.
The range of $g$, $\{m: \exists n(g(n)=m)\}$.
The first two are really the same thing in reverse mathematics: in (classical) reverse mathematics, we only have numbers and sets of numbers, so a "function of numbers" is really just a set with certain properties (namely, for each $n$ there is exactly one $k$ with $\langle n,k\rangle$ in the set).
The first two are also essentially the same in computability theory, in the sense that $g$ is computable iff its graph is (more generally, any function is of the same Turing degree as its graph).
The third, however, is a very different object (and in particular your claim "My question could be restated: "How can $\{\langle n,g(n)\rangle:n\in\mathbb{N}\}$ be computable if $g(n)$ itself isn't?"" is incorrect, since $\{\langle n,g(n)\rangle:n\in\mathbb{N}\}$ is very much not $\{m: \exists n(g(n)=m)\}$):
Let me first give an intuitive argument that the range of a computable function need not be computable, and then talk about what's going on in RCA$_0$.
Suppose you had a "black box" representing a function $g$, where you feed in natural numbers and it spits out natural numbers. Now you ask whether $3$ is in the range of (the function computed by) this black box. So you check:
Is $g(0)=3$? No - when you feed $0$ into the box, it spits out $2$, and $2\not=3$.
Is $g(1)=3$? No - when you feed $1$ into the box, it spits out $4$, and $4\not=3$.
Is $g(2)=3$? No - when you feed $2$ into the box, it spits out $17$, and $17\not=3$.
Is $g(3)=3$? No - when you feed $3$ into the box, it spits out $0$, and $0\not=3$.
...
However, at no point in this process can you confidently conclude that $3$ is not in the range of $g$! It could be that you just haven't tested the right value.
So even though you may think of the range of a function as having "less information" than the whole function itself - after all, which numbers go where is forgotten when we look at the range alone - this is actually horribly misleading in the context of computability theory and reverse math: extracting the range of a function from the function itself takes in general nontrivial power.
OK, so now let's think about RCA$_0$. You write:
This is not correct, and is also too vague; especially at first, you need to argue very precisely when trying to show that RCA$_0$ proves (or doesn't prove) something. RCA$_0$ has a very limited amount of comprehension (= set existence) power. In order to prove in RCA$_0$ that a set with a certain property exists, you basically need to show that that property can be expressed in a "simple" way - namely, in a $\Delta^0_1$ way. The obvious definition of the range of a function $g$ is $\Sigma^0_1$ with a parameter for $g$, so even if $g$ is computable we can't immediately argue that its range exists: we would need to also find a $\Pi^0_1$ definition for its range. And the point of the above section is that in general, we can't do better than $\Sigma^0_1$.
Now you might argue that in some informal sense, we can "construct" the range of a function from the function itself. That's reasonable, but then the way you're using the word "construct" is quite different from the way RCA$_0$ works, and the corresponding intuition will in fact be completely misleading in reverse mathematics.