Why can high variance help you in Heckman (1998) but not Aigner and Cain (1977)?
Say you have two distributions, $G$ and $H$ with equal means supported on (say) $[0,1]$ and $H$ is second-order stochastically dominated by $G$. Then, for any concave increasing $u$, we'll have $$\int udG \geq \int udH$$ This would correspond to Aigner and Cain, where the risk averse firm would prefer to hire an agent from distribution G. 

However, now suppose that $G$ and $H$ are distributions of talent (with the same stochastic ordering i.e. $H \succ_{C} G$). One equivalent notion to this idea is that distribution $G$ can be obtained by ``fusing" together part of the mass of $H$, i.e. by collapsing part of the mass to its respective barycenter. Given this, it is easy to see that there is always a cutoff $\eta \in [0,1]$ such that $H(\eta) \geq G(\eta)$. 

This corresponds to Heckman's result--if the job is sufficiently demanding, then it will be (at least weakly better) to be part of the high variance population. Moreover, note that this by no means depends on the expected quality of the agent. There are simply more of the highest quality people in the group with distribution $H$ than in the group with distribution $G$. They are not just more likely to exceed the cutoff, there are more of them that exceed the cutoff.

Another way to think about the hiring problem is as follows: say a firm wants to hire one person but receives applications from $10$ applicants, $5$ from group $A$ and $5$ from group $B$. A candidate's quality is unknown and it it ranges from $[0,1]$ continuously. Suppose also for the sake of simplicity that the firm may observe the candidate's group (and so knows what distribution he/she is drawn from): the distribution of quality for group $A$ is given by $G$ and the distribution of quality for group $B$ is given by $H$. Interviewing a candidate fully reveals a candidate's type, but the firm has the budget to interview only $4$ candidates. Whom should they interview?

The answer is simple: all four candidates interviewed should be from group $B$ (which has distribution $H$). Why? Because we can think about a candidate's quality as a random variable $X_{i}$ and thus the firm wants to maximize $\mathbb{E}[Z]$ where $$Z:=\max\big\{X_{1},X_{2},X_{3},X_{4}\big\}$$ Since $H \succ_{C} G$, $$\mathbb{E}_{H^{4}}[Z] \geq \mathbb{E}_{G^{4}}[Z]$$ 

Note also that $H \succ_{C} G$ $\Rightarrow$ $Var(H) \geq Var(G)$ but not $\Leftarrow$. In this sort of thing, I think it is stochastic dominance and not variance that is the apposite metric.