How was SHA-0 torn? - What is the meaning of a simple number of hash collisions?

I wanted to understand how the SHA0 hash function was violated. I understand that using hash problems / the principle of holding pigeons, hash collisions were found. http://www.mail-archive.com/cryptography%40metzdowd.com/msg02554.html contains an example message.

I'm having trouble finding / understanding: Does this mean that there is a timely, mathematical way to ALWAYS produce a hash collision?

Can I find m2 for a given m1, so that m1! = M2, sha (m1) == sha (m2), or is it possible only on a subset of the possible messages? Paraphrased: Is it likely that my password will receive another message per collision?

What is the meaning of searching for 2 random long messages, for example, in the link above that have the same hash value? Why should they sift through long random messages for a collision instead of calculating a collision for a practical message such as a “brown dog jumped over a fox”?

A few examples of hash collisions do not seem as important as the timely collision generation method for any message , but all messages speak of the first.

Thanks for any help / your time! I read a lot of posts / articles, but can't work on my confusion. I suspect I have the same questions for other broken hash functions like MD5.

EDIT:

The document (explaining the improved collision search method) referenced in the response

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Wikipedia:

2005 Xiaoyun , , SHA-0 2 ^ 39 .

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