...interpreting this as just a code snippet that focuses on your question,
e is a variable of type TimeoutException being specified as a parameter to your catch clause. If code running earlier in a "try block" happened to take a TimeoutException, a TimeoutException object would be instantiated, initialized and the catch() handler would receive that object as "e". The single line in the handler, e.printStackTrace(), calls the method printStackTrace() in the exception object. This causes the object to print a stack trace ( a way to "walk back" through the method calls that got you here... ) to the console. According to the API, it prints this stack trace to standard out.
SCJP - 86% - June 11, 2009
Well, sort of, yes. A catch block looks a bit like a little method; it has a parameter which it catches, and yes, you are applying the name "e" to any TimeoutExceptions you might catch. The printStackTrace() method is about the simplest action you can take in response to an Exception; it usually prints to the "standard error stream" (or wherever System.err happens to be pointing). It is actually inherited from the Throwable class. See the API specification for Exception (you can easily find a link to Throwable at the top of the web page), and the Java Tutorial about Exceptions.
I would warn you however that different people have different opinions about Exceptions.