$\newcommand\ep{\epsilon}$We have 1)$\implies$2) but 2)$\kern5pt\not\kern-5pt\implies$1). 

Indeed, suppose that 1) holds. Then, taking $f(x)\equiv x$ and letting 
$$Y_n:=X_n-X,$$
we have  
$$\sum_n E|Y_n|<\infty.$$
So, by Markov's inequality, for any real $\ep>0$ 
$$\sum_n P(|Y_n|\le\ep)=\sum_n(1-P(|Y_n|>\ep))
\ge\sum_n(1-E|Y_n|/\ep)=\sum_n 1-\sum_n E|Y_n|/\ep=\infty,$$
so that 2) holds. Thus, 1)$\implies$2). 

Now, as suggested in the comment by Martin Hairer, suppose that $X=0$ and $P(X_n=0)=P(X_n=1)=1/2$ for all $n$. Then $P(|Y_n|\le\ep)\ge1/2$ for all $n$ and hence 2) holds. On the other hand, $E|Y_n|=1/2$ and hence 1) does not hold (for $f(x)\equiv x$). Thus, 2)$\kern5pt\not\kern-5pt\implies$1).