This equivalence does not appear to be true in general, without further assumptions on $X$. Hausdorff is not enough, though regularity will suffice even if $X$ is not Hausdorff.
Counterexample.
[updated with $T_2$ counter-example]
The counterexample is essentially that of this answer from a post linked by Ulli in the comments. Concretely, let $X=[0,1]$ with the $K$-topology given by open sets of the form $U\backslash S$, where $U$ is open in the usual topology and $S\subseteq K:=\{\frac{1}{n}\mid n\in \mathbb N\}$. Note that as this topology is stronger than the standard one, it is Hausdorff.
It is easy to verify that for $A\subseteq X$, conditions 4, 5, and 6 are equivalent to each other, and to the condition that $A\cap K$ is finite. Moreover, since $X$ is first countable (in fact second countable), every subset $A$ satisfies condition 2 in definition 2, hence $X$ is Angelic.
However, conditions 1, 2, and 3 are equivalent to the stronger condition that the closure of $A$ (in either the usual or $K$-topology) contains only finitely many members of $K$.
Regularity is sufficient.
If a (not necessarily Hausdorff) space $X$ is regular, than regardless of whether or not it is angelic, we have that $N$-compactness (condition 4) implies relative compactness (condition 1), so that for angelic spaces in particular the equivalence is closed.
To see this, if $A\subseteq X$ is not relatively compact, then let $\mathcal U$ be an open cover of $\overline{A}$ with no finite subcover. Then we let $\mathcal F$ be the family of all closed subsets $F\subseteq X$ such that $F\subseteq U_1\cup\dots\cup U_n$, for some finite subfamily $\{U_i\}\subset \mathcal U$. Note that $\mathcal F$ is directed by inclusion.
For each $F\in \mathcal F$, we may choose a point $x_F\in A\backslash F$, as such a point exists by the density of $A$ in $\overline{A}$, and the fact that $\overline{A}\backslash F\neq \emptyset$.
Then for every $x\in \overline{A}$, there is some $U\in\mathcal U$ with $x\in U$, and by regularity, some open $V$ with $x\in V\subseteq \overline{V}\subseteq U$, and so eventually the net $\{x_F\}$ will lie outside of $V$, since $\overline{V}\in\mathcal F$, hence no subnet can converge to $x$. Since this holds for all $x\in \overline{A}$, we have a net in $A$ with no subnet converging in $\overline{A}$, so $N$-compactness fails.
Remark.
In general, it seems $N$-compactness is indeed equivalent to Definition 3 in the aforementioned post linked by Ulli.
Definition 3. $A$ is relatively compact in $X$ if every open cover of $X$ has a finite subset that covers $A$...
(Incidentally, this means earlier we could have appealed to a different answer from that same question to show that under regularity, all definitions are equivalent.)
The proof is straightforward:
If $A$ fails Definition 3, let $\mathcal U$ be an open cover of $X$ with no finite subset covering $A$, and consider the directed system $\mathcal V$ of finite unions $V=U_1\cup\dots\cup U_n$ with each $U_i\in\mathcal U$; then as in the preceding argument construct a net of points in $A$ that is eventually outside of each $V\in \mathcal V$, hence a fortiori outside of each $U\in\mathcal U$, which therefore cannot converge in $X$.
Conversely, if $A$ satisfies Definition 3, and $\{x_i\mid i\in I\}$ is a net in $A$, then argue there is some $x\in X$ for which $x_i$ frequently lies in every neighborhood (otherwise we get a contradiction to Definition 3). Then construct a convergent subnet in the usual way as done here, for example. Of course as a limit of a net in $A$, we have $x\in \overline{A}$.
