What does "typedef void (*Something)()" mean

It defines a pointer-to-function type. The functions return void, and the argument list is unspecified because the question is (currently, but possibly erroneously) tagged C; if it were tagged C++, then the function would take no arguments at all. To make it a function that takes no arguments (in C), you'd use:

typedef void (*MCB)(void);

This is one of the areas where there is a significant difference between C, which does not - yet - require all functions to be prototyped before being defined or used, and C++, which does.

It introduces a function pointer type, pointing to a function returning nothing (void), not taking any parameters and naming the new type MCB.

The typedef defines MCB as the type of a pointer to a function that takes no arguments, and returns void.

Note that MCB Modes::m_process = NULL; is C++, not C. Also, in C, the typedef should really be typedef void (*MCB)(void);.

I'm not sure what you mean by "the memory was freed". You have a static pointer to a function; a function cannot be freed. At most, your pointer has been reset somewhere. Just debug with a memory watch on m_process.