Signature and hash function properties I understand the need for the hash function to be collision resistant and second pre-image resistant. For what reason, exactly, does a hash function need to be pre-image resistant?

If this property is not useful for signatures, what is the origin of this requirement, maybe privacy?
The question "why is preimage resistance needed for hash functions" is not really relevant. This is because collision resistance implies preimage resistance. Thus, it is just a fact that if you have collision resistance then you must have preimage resistance.

So, instead, I will relate to what preimage resistance is good for at all. In more technical cryptographic terms, a preimage resistant function is called a one-way function. One-way functions are the most basic cryptographic primitive and you can construct all of symmetric crypto from them (i.e., you can construct pseudorandom generators, functions, symmetric encryption, message authentication codes, and so on). [Note: you cannot construct collision-resistant hash functions from them in a black-box way.] Thus, from a theoretical perspective, these functions are very interesting; I say "theoretical" since the above constructions are all theoretical and not practical.

Note that preimage resistance is a necessary condition almost everywhere in cryptography, but it is usually not sufficient. This is because it does not mean that it's impossible to obtain half of the preimage, and it is also only meaningful for very high entropy - if not random - inputs. The use of preimage resistance in hashing passwords is used in practice, but is only heuristic. In order to analyze this properly, you actually need to model the function as a random oracle.

So, preimage resistance or one-wayness is a fact of cryptography. It is the minimal property that you need to do almost anything interesting in crypto (apart from all of the information theoretic crypto work which I won't discuss here). However, it is not a security notion that is usually of interest in and by itself.

A proof sketch that collision resistance implies one-wayness

Assume that there exists an adversary $A$ who can invert a function $H:\{0,1\}^*\rightarrow\{0,1\}^n$ with probability $\epsilon$ on a random input of length $2n$. We construct an adversary $A'$ who finds a collision in $H$. Adversary $A'$ chooses a random $r\in\{0,1\}^{2n}$, computes $y=H(r)$ and invokes $A$ on input $y$. If $A$ returns $s$ such that $H(s)=y$ and $s\neq r$ then $A'$ outputs $s$. Otherwise it outputs $\bot$. We now analyze the probability that $A'$ succeeds. We have that $A'$ succeeds if $A$ succeeds and $s\neq r$. Since the input is of length $2n$ and the output is of length $n$, we have that each output has on average $2^n$ premiages. Thus, the probability that $s=r$ is negligible. [This part of the proof needs some work, but I'll leave that to the readers as an exercise.] We therefore conclude that $A'$ succeeds with probability $\epsilon - neg(n)$ where $neg$ is a negligible function. Thus, if $\epsilon$ is non-negligible then the function $H$ is not collision resistant. We conclude that if $H$ is collision resistant, then $A$ could only succeed with negligible probability and so $H$ is one-way.