Xor'ing sha256 words instead of adding In sha-256, when the words go through the compression function, they get added to the previous hash. In case of the first 64 rounds, they get added to the initial words ($\bmod 2^{32}$ of course). However, would xor'ing instead of adding be more secure, or not?
Any group operation $\odot$ on the set $\{0,1\}^{256}$ equally matches the objective of making the hash round function $F(m,x)=E(m,x)\odot x$ one-way under the assumption that $x\mapsto E(m,x)$ is a random permutation keyed by $m$. Thus under this model, XOR or 32-bit addition does not matter.


  would xor'ing instead of adding be more secure, or not?


For the reason above: That's moot. But if we need to choose, rather not. Argument: addition makes the natural expression of a hash round function as a number of elementary logic gates (or satisfiability equations) slightly more complex than XOR due to the  8×31 carry terms. And absent any other good criteria, it's reasonable to bet that more complex is more secure, because that's demonstrably true for a random arrangement of gates (or satisfiability equations)