Call the set $S_k = \{1, 2, \dots, k\}$ a block, for $k=1, 2, \dots$; for consistency assume that $S_0$ is the empty block. Suppose that several numbers can be partitioned into blocks. The key observation is that the same holds true after any operation is applied. Indeed let $S_k$ be one of the blocks, and let $f(a)$ denote the operation applied to number $a$. If $a > k$ then $S_k$ remains unchanged. If $a \le k$ then $S_k$ is replaced by the blocks $S_{a-1}$ and $S_{k-a}$. The claim is justified.

Since the initial numbers $1, 2, \ldots, 2013$ form a block, it follows that one has a disjoint union of blocks after any number of operations. The sum of the block $S_k$ is $\frac{1}{2}k(k+1)$, so if the total sum is $10$ then the blocks participating can be only $S_1 = \{1\}$, $S_2 = \{1, 2\}$, $S_3 = \{1, 2, 3\}$, $S_4 = \{1, 2, 3, 4\}$, with sums $1, 3, 6, 10$ respectively. So the question reduces to representing $10$ as a sum of several numbers among $1, 3, 6, 10$. The possibilities are