"error: assignment to expression with array type error" when I assign a struct field (C)

You are facing issue in

 s1.name="Paolo";

because, in the LHS, you're using an array type, which is not assignable.

To elaborate, from C11, chapter §6.5.16

assignment operator shall have a modifiable lvalue as its left operand.

and, regarding the modifiable lvalue, from chapter §6.3.2.1

A modifiable lvalue is an lvalue that does not have array type, [...]

You need to use strcpy() to copy into the array.

That said, data s1 = {"Paolo", "Rossi", 19}; works fine, because this is not a direct assignment involving assignment operator. There we're using a brace-enclosed initializer list to provide the initial values of the object. That follows the law of initialization, as mentioned in chapter §6.7.9

Each brace-enclosed initializer list has an associated current object. When no designations are present, subobjects of the current object are initialized in order according to the type of the current object: array elements in increasing subscript order, structure members in declaration order, and the first named member of a union.[....]

typedef struct{
     char name[30];
     char surname[30];
     int age;
} data;

defines that data should be a block of memory that fits 60 chars plus 4 for the int (see note)

[----------------------------,------------------------------,----]
 ^ this is name              ^ this is surname              ^ this is age

This allocates the memory on the stack.

data s1;

Assignments just copies numbers, sometimes pointers.

This fails

s1.name = "Paulo";

because the compiler knows that s1.name is the start of a struct 64 bytes long, and "Paulo" is a char[] 6 bytes long (6 because of the trailing \0 in C strings)
Thus, trying to assign a pointer to a string into a string.

To copy "Paulo" into the struct at the point name and "Rossi" into the struct at point surname.

memcpy(s1.name,    "Paulo", 6);
memcpy(s1.surname, "Rossi", 6);
s1.age = 1;

You end up with

[Paulo0----------------------,Rossi0-------------------------,0001]

strcpy does the same thing but it knows about \0 termination so does not need the length hardcoded.

Alternatively you can define a struct which points to char arrays of any length.

typedef struct {
  char *name;
  char *surname;
  int age;
} data;

This will create

[----,----,----]

This will now work because you are filling the struct with pointers.

s1.name = "Paulo";
s1.surname = "Rossi";
s1.age = 1;

Something like this

[---4,--10,---1]

Where 4 and 10 are pointers.

Note: the ints and pointers can be different sizes, the sizes 4 above are 32bit as an example.

Please check this example here: Accessing Structure Members

There is explained that the right way to do it is like this:

strcpy(s1.name , "Egzona");
printf( "Name : %s\n", s1.name);