‘\n\t\r’ == 0 is true?

This behaviour might be surprising but can be explained by having a look at the specification.

We have to look at the what happens when a comparison with the equals operator is performed. The exact algorithm is defined in section 11.9.3.

I built a simple tool to demonstrate which algorithm steps are executed: https://felix-kling.de/js-loose-comparison/

string == integer

The step we have to look at is #5:

5. If Type(x) is String and Type(y) is Number,
return the result of the comparison ToNumber(x) == y.

That means the string "\n" ("\r", "\t") is converted to a number first and then compared against 0.

How is a string converted to a number? This is explained in section 9.3.1. In short, we have:

The MV (mathematical value) of StringNumericLiteral ::: StrWhiteSpace is 0.

where StrWhiteSpace is defined as

StrWhiteSpace :::
    StrWhiteSpaceChar StrWhiteSpace_opt

StrWhiteSpaceChar :::
    WhiteSpace
    LineTerminator

This just means that the numerical value of strings containing white space characters and/or a line terminator is 0.
Which characters are considered as white space characters is defined in section 7.3.

string == boolean

The step we have to look at is #7:

7. If Type(y) is Boolean, return the result of the comparison x == ToNumber(y).

How booleans are converted to numbers is pretty simple: true becomes 1 and false becomes 0.

Afterwards we are comparing a string against a number, which is explained above.

As others have mentioned, strict comparison (===) can be used to avoid this “problem”. Actually you should only be using the normal comparison if you know what you are doing and want this behaviour.