Accessing arrays in an array of pointers

Yes, you will (assuming that the array has been filled correctly). Imagine the double pointer situation as a table. You then have the following, where each string is at a completely different memory address. Please note that all addresses have been made up, and probably won't be real in any system.

strings[0] = 0x1000000
strings[1] = 0xF0;
...
strings[n] = 0x5607;

0x1000000 -> "Hello"
0xF0 -> "World"

Note here that none of the actual text is stored in the strings. The storage at those addresses will contain the actual text though.

For this reason, strings + 2 will add two to the strings pointer, which will yield strings[2], which will yield a memory address, which can then be used to access the string.

strings + 2 is the address of the 3rd element of the buffer pointed to by string.
*(strings + 2) or strings[2] is the 3rd element which is again a pointer to a buffer of characters.

i think you are looking to access third element through expression

strings[2];

but this will not be the case because look at the type of expression string[2]

Type is char *

As according to the standards

A 'n' element array of type 't' will be decayed into pointer of type __t__.With the exception when expression is an operand to '&' operator and 'sizeof' operator.

so strings[2] is equivalent to *(strings + 2) so it will print the contents of the pointer to pointer at third location,which is the contents of a pointer i.e an address.
But

strings+2;

whose type is char ** will print the 3 rd location's address,i.e,address of the 3rd element of array of pointer, whose base address is stored in **string.

But in your question you have not shown any assignment to the char ** strings and i am answering by assuming it to be initialised with particular array of pointers.
According to your question it is silly to do

*(strings + 2)

As it is not initialised.