I think the issue is understanding how data is actually stored in memory, as opposed to how it's abstracted when using a high-level language. From your confusion over how the string is stored in memory, I think that's where you should start.
At the level you're operating on in MIPS, there is no "string" (only Zuul). There's not really an "int" or a "float" in the way you're used to thinking of them, either. There's just bits (1/0) which are grouped together in standard ways to compose the data types you're used to using. For reasons your professor will be telling you about, we use groups of 8 bits, which we call a "byte".
Each of those characters in your string can be represented in a byte. Take a look at http://www.asciitable.com/ to see the standard for how that is done. So the 'L' is represented by a string of bits that corresponds to the hex value 0x4c, which just happens to be decimal value 76.
That value is stored in a single byte in RAM. The next letter ('o', represented by hex value 0x6f) is stored in the next sequential byte. The next letter is stored in the next byte, all the way through the last one. If you had to come up with a name for that chain of sequential bytes, you might call it a "string" of characters. :)
As @Michael said, you need to know that a zero-terminated ASCII string (also called an ASCIIZ string) has an additional character at the end. That character is represented by hex value 0x00 (plain old zero), and you'll see it referenced as '\0'.
One other thing. You need to find a better ASCII-to-Hex translation tool. The one you used didn't treat the "space" character correctly. A space corresponds to ASCII character 0x20 (decimal 32). If you leave it out, your string is "LongNumbers". If you try putting 0x00 in there, your string will end after "Long".