When I run these lines in my terminal on macOS I get the correct result
hex=$(echo -n 'betty' | xxd -p) echo $hex 6265747479 but when I run them in a bash script I get something completely different
sh myscript.sh 62657474790a It's like its put a carriage return on the end for some reason. WHY?
Different versions of echo do different things when given -n as their first argument. Some print it as part of their output, some interpret it as a flag, who knows what's going to happen. According to the POSIX standard for echo, "If the first operand is -n, or if any of the operands contain a backslash character, the results are implementation-defined."
The most reliable way to print a string without a linefeed after it is with printf. It's slightly more complicated because you have to give it a format string as well as the string you want printed:
hex=$(printf "%s" 'betty' | xxd -p) 
Try bash instead of sh. As echo is a builtin command, it depends on the shell how it is executed. For me, the echo from sh does not seem to support -n:
sh-3.2$ echo -n hello -n hello sh-3.2$ echo $(echo -n 'betty' | xxd -p) 2d6e2062657474790a 
cat myscript.sh; sh myscript.shshas a shell, and we don't know where sh is linked to on your machine. If both are bash shells, they might be different versions (do aprint $BASH_VERSIONin both variants), which implement theechoin a different way.echo -n 'betty'would produce that output, which appears to bebettyfollowed by a newline. I'd expect eitherbettywithout a newline, or-n bettywith a newline. What exactly doessh -c 'echo -n betty'print, compared toecho -n betty? What doesls -l /bin/shsay?sh -c 'echo -n betty'gives me-n bettyecho -n bettygives mebettywith the next prompt on the same line as bettyls -l /bin/sh=>-r-xr-xr-x 1 root wheel 630464 29 Apr 10:31 /bin/sh