variables f,g,h,i and j are assigned to registers $s0, $s1, $s2, $s3, and $s4 respectively

The base addresses for the arrays A and B are in registers $s6 and $s7

addi $t0, $s6, 4    #load the forth element of array A into $t0
add  $t1, $s6, $0   #load the first element of array A into $t1
sw   $t1, 0($t0)    #store $t0(the forth element of array A) into $t1
lw   $t0, 0($t0)    #load $t0 into $t0
add  $s0, $t1, $t0  #add $t1 and $t0 into $s0 

comments are added by me and are my attempt at understanding the code

I really don't understand lines three and four. How did these temporary registers "become arrays"?
specifically the part '0($t0)' is what's confusing me. Wouldn't this imply that $t0 is an array and we're trying to access the first element of that array?

P.S. I have to convert this code to C after, so I'll post my conversion after I've gained a better understanding of this code.

The first two lines don't load anything. You'd need a load instruction for that. They simply perform pointer arithmetic.

The first line makes $t0 point to the fifth byte of the A array. If A is an array of 32-bit values, that means that $t0 points to the second element of the array - not the fourth or fifth.

The third line takes the value that's in the register $t1 (i.e. the address of A) and stores it in the memory location that $t0 points to, i.e. it stores it as the first element of A.

The fourth lines takes the second element of A (which has been set on the previous line to be the address of A) and stores it in the register $t0 (which previously stored the address of that element).

You seem to understand the fifth line fine.

Since the same question has been asked on dreamincode, I'll stop responding here. There's no point in having the same discussion twice.