Try reading it piece by piece. Recall that $A\cup B$ means that at least one of $A$, $B$ happens and $A\cap B$ means that both $A$ and $B$ happen. Infinite unions and intersections are interpreted similarly. In your case, $\bigcup_{k=n}^{\infty}A_k$ means that at least one of the events $A_k$ for $k\geq n$ happens. In other words "there exists $k\geq n$ such that $A_k$ happens".
Now, let $B_n=\bigcup_{k=n}^{\infty}A_k$ to simplify notation a bit. This gives us $\bigcap_{n=0}^{\infty}\bigcup_{k=n}^{\infty}A_k = \bigcap_{n=0}^{\infty}B_n$. This is interpreted as "all of the events $B_n$ for $n\geq 0$ happen" which is the same as "for each $n\geq 0$ the event $B_n$ happens". Combined with the above interpretation, this tells us that that $\limsup A_n$ means "for each $n\geq 0$ it happens that there is a $k\geq n$ such that $A_k$ happens". This is precisely the same as saying that infinitely many of the events $A_k$ happen.
The other one is interpreted similarly: $\bigcap_{k=n}^{\infty}A_k$ means that for all $k\geq n$ the event $A_k$ happens. So, $\bigcup_{n=0}^{\infty}\bigcap_{k=n}^{\infty}A_k$ says that for at least one $n\geq0$ the event $\bigcap_{k=n}^{\infty}A_k$ will happen, i.e.: there is a $n\geq 0$ such that for all $k\geq n$ the event $A_k$ happens. In other words: $\liminf A_n$ is the event that from some point on, every event happens.
Edit: As requested by Diego, I'm adding a further explanation. Sets are naturally ordered by inclusion $\subseteq$. This is a partial order, even a lattice. (Putting aside the fact that the universe of sets is not a set.) In fact, every family of sets has an $\inf$ and $\sup$ with respect to $\subseteq$, which can be defined by: $$\inf_{\lambda\in\Lambda}A_\lambda =\bigcap_{\lambda\in\Lambda}A_\lambda$$ and $$\sup_{\lambda\in\Lambda}A_\lambda =\bigcup_{\lambda\in\Lambda}A_\lambda.$$
Now, the usual definition of $\limsup$ and $\liminf$ (of sequences of real numbers) can be rephrased in terms of infima and suprema as follows: $$\liminf_{n\to\infty}a_n=\sup_{n\geq 0}\inf_{k\geq n} a_n$$ and $$\limsup_{n\to\infty}a_n=\inf_{n\geq 0}\sup_{k\geq n} a_n.$$
We can now use the same definition for sets: $$\liminf_{n\to\infty}A_n=\sup_{n\geq 0}\inf_{k\geq n} A_n$$ and $$\limsup_{n\to\infty}A_n=\inf_{n\geq 0}\sup_{k\geq n} A_n.$$ Rewriting this in terms of $\bigcup$ and $\bigcap$, we get precisely the definitions from the question.
Best Answer
The first definition is easiest to understand in my opinion. It describes all those sets for which only finitely many $A_n$ are not in $F$.
You can see that the first definition for "almost always in $F$" is equal to $\liminf_{n\rightarrow \infty}A_n$. As pointed out in the comments $B_n = \bigcap_{j \geq n} A_j$ is an increasing sequence and thus $$\lim_{n\rightarrow \infty} B_n = \liminf_{n\rightarrow \infty}B_n = \bigcup_{n \geq 1} \bigcap_{j \geq n} B_j = \bigcup_{n \geq 1} B_n = \bigcup_{n \geq 1} \bigcap_{j \geq n} A_j= \liminf_{n\rightarrow \infty}A_n.$$