I'm reading through a proof of Lindstrom's theorem (If a logic stronger than first-order logic satisfies Lowenheim-Skolem and Compactness then it is equivalent to first order logic) but their formulation of "this logic satisfies Lowenheim-Skolem" is a bit different to the Lowenheim-Skolem theore I'm used to.
I include here both formulations:
No, they are not equivalent. The property $LoSko(\mathcal{L})$ is the weakest version of the Lowenheim-Skolem property for first-order logic that is considered (at least as far as I'm aware); it's used because it leads to the strongest version of Lindstrom's theorem (since it yields a weaker hypothesis).
(Incidentally, working with the slightly stronger version $LoSko^+$ (see below) makes the proof of Lindstrom's theorem meaningfully simpler, since you don't have to worry about expressing the relevant theory in a single sentence. If you're presenting the theorem, I recommend presenting it this way and then mentioning at the end how it can be improved.)
In fact, the following property is strictly stronger than $LoSko$:
To produce a logic satisfying $LoSko$ but not $LoSko^+$, we obviously need to work in a logic without infinitary conjunctions, so infinitary logic is right out. And second-order logic doesn't even satisfy $LoSko$, so that's out too. In fact, no natural logic that I'm aware of provides a counterexample. However, unnatural counterexamples can easily be built.
One way to do this is as follows. Fix a complete first-order theory $T$ with countably infinitely many countable models $\{M_i: i\in\mathbb{N}\}$ up to isomorphism. Let $\mathcal{L}$ be the smallest regular logic containing first-order logic and sentences $\psi_i$ saying "I am not $M_i$" for each $i\in\mathbb{N}$. It's not hard to show that this logic satisfies $LoSko$, essentially because adding few sentences picking out exactly one structure doesn't change very much, other than absolutely destroying compactness. But clearly $\mathcal{L}$ doesn't have $LoSko^+$, since $T\cup\{\psi_i: i\in\mathbb{N}\}$ is satisfiable but has no countable models.