Stdout string manipulate, delete all after a word per line

This is basic sed. Using sed is not difficult once you know regular expressions. A basic sed command for reading the input and stripping every < and the following part if it exist, then printing the line (may be modified):

$ echo 'foo<....' | sed 's/<.*//'
foo

sed uses regular expressions, the relevant manual page text for sed(1) that applies to the above command:

s/regexp/replacement/
Attempt to match regexp against the pattern space. If successful, replace that portion matched with replacement. The replacement may contain the special character & to refer to that portion of the pattern space which matched, and the special escapes \1 through \9 to refer to the corresponding matching sub-expressions in the regexp

Alternative using cut (manual page for cut(1)), "split the string by < and take the 1st field.

echo 'foo<....' | cut -d'<' -f1

Alternative using grep, "match only everything containing characters of the set a to z (case insensitive)" (manual page of grep(1)):

echo 'foo<....' | grep -io '[a-z]*'

(note: I took the liberty to use [a-z]*, meaning "zero or more occurrences of a letter", because grep won't return an empty line when using the -o option)

Alternative using awk, using the same idea of cut (manual page of awk(1)):

echo 'foo<....' | awk -F '<' '{print $1}'

Sample:

cat textWithFooBar.txt | cut -d '<' -f 1 > output.txt