What does it mean to be simultaneously hardcore?
Hard-core bits are related to one-way functions.

For some intuition on what hard-core bits are, consider a one-way function $f$. SInce it's a one-way function, it's hard to invert: that is, if I select a random $x$ in the function's domain and give you $f(x)$, you cannot find a $x'$ such that $f(x) = f(x')$ with non-negligible probability in probabilistic polynomial time (this is the definition of a one-way function). 

So, intuitively, there is "something" about $x$ that is hidden from you in $f(x)$. If you knew everything about $x$ from $f(x)$, you would know what $x$ was. So, there is something that $f$ hides about its input. That something is a "hard-core predicate". 

A hard-core predicate is itself a function. Let's say we have a hard-core predicate for $f$ called $\mathsf{hc}$. Since it's a predicate, $\mathsf{hc}$ returns exactly one bit.  So, given $f(x)$, you cannot determine the value of $\mathsf{hc}(x)$ with non-negligible probability (over $1/2$, which you can always get by guessing) in probabilistic polynomial time. In essence, $\mathsf{hc}(x)$ represents something about $x$ that you do not know given $f(x)$ &mdash; that is, $\mathsf{hc}$ is a bit related to the input that $f$ hides from you. Another way of looking at it is that $\mathsf{hc}(x)$ is indistinguishable from random (in PPT and all that business). To see this, consider if you were given $f(x)$ and $\mathsf{hc}(x)$. Since you cannot compute $\mathsf{hc}(x)$ from $f(x)$, by definition, you should not be able to distinguish it from a random bit.

Note that a hard-core predicate is a single bit. Instead of a single bit, the paper wishes to talk about multiple different bits each being hard-core simultaneously. So, simultaneously-hardcore bits are a natural extension of hard-core predicates into strings that are multiple bits in length. To be explicit, let $h$ be a hard-core function for $f$. Then, given $f(x)$ for a random $x$ in $f$'s domain, an adversary cannot compute $h(x)$ in PPT. The paper you've linked frames $h$ as being indistinguishable from random, which is equivalent.