What gives SHA-256 its preimage resistance?
You appear to be thinking "the output of the round function appears to be a simple function of the previous round, and a word from the message; why can't we select message words to steer the output to whatever we wanted".

Well, there's something missing in the simple description you gave; message expansion.  The W inputs are not just directly taken from the input message; the first 16 are, but the other 48 are linear functions of the input message.  If you modify one input word, you necessarily modify a number of the W inputs; this effectively stops this simple steering strategy, as tweaking one W input (to get a desired state) will necessarily modify others (which will modify the state you're trying to set up).

This doesn't prove the SHA-256 is secure (SHA-1 does the same sort of message expansion, and we know that's weak); however it does show that this approach doesn't appear to be fruitful.