When something like this happens, take your expression apart. Look at the pieces. Then think about what MATLAB is telling you.
location = 'nswe';
strfind(location,'o') > 0
strfind(location,'e') > 0
So the first fragment returns empty, the second returns true.
What does the operator do there?
 || 1
Operands to the || and && operators must be convertible to logical scalar values.
Is an empty result a scalar, logical value? No. So you need to think about how to better write that test to not fail.
For example, since you know that strfind may return zero, one, or more than one solution depending on the string, you need to write code that will succeed in any case. Of course, I'm not terribly sure why you are testing to see if 'o' falls in the string 'nswe', but maybe you have a valid reason.
(I could show you a better way to write that test, but I don't know what strings you might have as possibilities. Only you know your real problem.)