Why do I need to use type** to point to type*?

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No, that tutorial is of questionable quality. I wouldn’t recommend to continue reading it.

A char** is a pointer-to-pointer. It is not a 2D array.
It is not a pointer to an array.
It is not a pointer to a 2D array.

The author of the tutorial is likely confused because there is a wide-spread bad and incorrect practice saying that you should allocate dynamic 2D arrays like this:

// BAD! Do not do like this!
int** heap_fiasco;
heap_fiasco = malloc(X * sizeof(int*));
for(int x=0; x<X; x++)
{
  heap_fiasco[x] = malloc(Y * sizeof(int));
}

This is however not a 2D array, it is a slow, fragmented lookup table allocated all over the heap. The syntax of accessing one item in the lookup table, heap_fiasco[x][y], looks just like array indexing syntax, so therefore a lot of people for some reason believe this is how you allocate 2D arrays.

The correct way to allocate a 2D array dynamically is:

// correct
int (*array2d)[Y] = malloc(sizeof(int[X][Y]));

You can tell that the first is not an array because if you do memcpy(heap_fiasco, heap_fiasco2, sizeof(int[X][Y])) the code will crash and burn. The items are not allocated in adjacent memory.

Similarly memcpy(heap_fiasco, heap_fiasco2, sizeof(*heap_fiasco)) will also crash and burn, but for other reasons: you get the size of a pointer not an array.

While memcpy(array2d, array2d_2, sizeof(*array2d)) will work, because it is a 2D array.

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