What makes SHA-2 and SHA-3 have security levels half their output hash length? The tag wiki for sha-256 says, in part:


  SHA-256 ... processes the input in 512-bit block and has a 256-bit output, for a 128-bit security level.


I'm not a cryptographer, just an interested amateur, but how does it follow that SHA-256 has only a 128-bit security level? (More generally about the whole SHA-2 and SHA-3 families.) Also, against what type of attack is this considered?

I looked to Wikipedia which didn't really help my confusion. The cryptanalytic results listed (up to 2014) are all for reduced-rounds variants, and most for SHA-256 are on the order of $2^{250}$ complexity or more. The comparison of SHA functions gives the security of each SHA-2 and SHA-3 family function as half the output hash length, but provides no clarification of how this figure is derived.

Is the security level related to the relationship between the input processing block size and the output hash length?

Please go easy on me with the math; plain English explanations backed up with some math are preferred.
You can't say SHA-2 has a security level of half its hash length without any given context.  128 bit against what type of attack?  What is the attacker trying to do?  Perform a collision? Ok yes it has 128 bit against collisions. Perform a preimage? Nope it has 256 bit security against preimage attacks.  

Any algorithm is considered n bit strength if the expected number of operations for the average case is 2^n.  For unbroken cryptographic hashes the strength is 2^(n/2) against collisions and 2^n against preimages.  Collisions are simply "easier" than preimage attacks due to the birthday problem.  They take less operations so you need a larger search space (digest length) to achieve the same level of security.

So the statement that SHA-256 has 128 bit security is incomplete and it ignores the fact that collisions are not always relevant.  Against many type of scenarios a collision is of no value.  An example where collisions are not a threat would be hashed passwords.  For an attacker to break a specific password they need to find a preimage of the password hash.  In that application SHA-256 has 256 bit security because collisions present no security risk(i.e. attacker determines abc and xyz both hash 1234 and that doesn't match any password hash).  It doesn't help that the word collision is used in common language to scenarios which refer to a preimage.  Also keep in mind there are two types of preimage attacks and once algorithms become partially compromised they may have differing levels of security depending on if it is a first or second preimage attack.  Of course once an algorithm is compromised you should look to migrate to a more secure algorithm but that distinction could be relevant in some situations.