I'm adding an answer to my own question to point out one thing about Kleene's $\mathcal{K}_2$ which I had previously failed to understand and which is an important difference with $\mathcal{K}_1$ that probably deserves its own special caveat in any description of $\mathcal{K}_2$: in $\mathcal{K}_1$ one can “run programs in parallel”, but apparently one cannot do this in $\mathcal{K}_2$.
That is, $\mathcal{K}_1$ has the property that given $e_1,e_2$ (in $\mathcal{K}_1$, that is to say, $\mathbb{N}$) there exists $e$ such that that $e\bullet n$ is defined iff either $e_1\bullet n$ or $e_2\bullet n$ is defined, and moreover, when this is the case, $e\bullet n$ equals one of the values $e_1\bullet n$ or $e_2\bullet n$ which is defined (so, the one which is defined if only one is defined). Furthermore, we can find such $e$ effectively from $e_1,e_2$ (meaning that there is a $p$ such that $(p\bullet e_1)\bullet e_2$ gives such an $e$ as I wrote). Indeed, we do this by running $e_1$ and $e_2$ in parallel until one produces a result, and returning the result in question (this is a consequence of Kleene's normal form theorem).
This property, however, does not hold for $\mathcal{K}_2$: while it is true that ① given $\varepsilon_1,\varepsilon_2$ (in $\mathcal{K}_2$, that is to say, $\mathbb{N}^{\mathbb{N}}$) there exists $\varepsilon$ such that that $\varepsilon\bullet \nu$ is defined iff either $\varepsilon_1\bullet \nu$ or $\varepsilon_2\bullet \nu$ is defined, however we cannot ② add the condition that, $\varepsilon\bullet \nu$ equals one of the values $\varepsilon_1\bullet \nu$ or $\varepsilon_2\bullet \nu$ which is defined, not even if we assume that $\varepsilon_1\bullet \nu$ and $\varepsilon_2\bullet \nu$ are never simultaneously defined.
Indeed, this is obvious topologically: as explained in Andrej Bauer's notes on realizability, theorem 2.1.14, the partial maps $\mathbb{N}^{\mathbb{N}} \dashrightarrow \mathbb{N}^{\mathbb{N}}$ of the form $\nu \mapsto \varepsilon\bullet\nu$ are precisely the continuous maps on a $G_\delta$ domain. Now $G_\delta$ sets are closed under finite union, so this shows ①; however, given two continuous functions on disjoint $G_\delta$ sets there does not necessarily exist one which extends them on the union (disproving ②). Indeed, a trivial counterexample is given by the partial function $F_1\colon \mathbb{N}^{\mathbb{N}} \dashrightarrow \mathbb{N}^{\mathbb{N}}$ taking the constant function with value $0$ at the constant function with value $0$ and undefined elsewhere, and the partial function $F_2\colon \mathbb{N}^{\mathbb{N}} \dashrightarrow \mathbb{N}^{\mathbb{N}}$ taking the constant function with value $1$ everywhere except at the constant function with value $0$ and undefined there. Clearly $F_1$ and $F_2$ are continuous and each defined on a $G_\delta$, but the unique function $F$ extending them to the union of their domains, i.e. to $\mathbb{N}^{\mathbb{N}}$, is not continuous, so not realized by an element of $\mathcal{K}_2^{\mathrm{eff}}$. Moreover, each of $F_1$ and $F_2$ belong to the effective part of $\mathcal{K}_2$ (the one realized by Turing-computable functions in $\mathbb{N}^{\mathbb{N}}$), so this defect(?) of $\mathcal{K}_2$ caries over to $\mathcal{K}_2^{\mathrm{eff}}$.
The difficulty arises from the way we move from functions $\mathbb{N}^{\mathbb{N}} \dashrightarrow \mathbb{N}$ to functions $\mathbb{N}^{\mathbb{N}} \dashrightarrow \mathbb{N}^{\mathbb{N}}$. The former have no such problem: the ones described by an element of $\mathbb{N}^{\mathbb{N}}$ are precisely the continuous functions on an open domain, and we can glue such functions (feed the input function and take the first one that returns a value), but when going to $\mathbb{N}^{\mathbb{N}} \dashrightarrow \mathbb{N}$ things break apart as the previous counterexample shows.
I don't know if this should be considered a “defect” of $\mathcal{K}_2$, but I urge anyone writing a explanation of $\mathcal{K}_2$ to add a warning to the reader that this property we tend to take for granted in $\mathcal{K}_1$ does not hold for $\mathcal{K}_2$ and it's likely to confuse beginners.