Is this a good way to (or correct) proof? 
Given: P ⇒ Q. To show: ¬Q ⇒ ¬P.

Using modus tollens:
Assume that ¬Q.
Then ¬P.
Thus ¬Q ⇒ ¬P.
Does anybody have a good reference to how I can learn how to proof such formulas (including equivalence, negation etc.)
Thank you!
 A: 
"Does anybody have a good reference to how I can learn how to prove such formulas?" 

There is no single formal proof system for propositional logic. There are different general styles of proof system (axiomatic, natural deduction, tableaux ...), and within each general style there are variations (there are, for example, different natural deduction systems with different basic rules of inference, and different ways of laying out proofs). It is important to realise this or you can get confused!
So: which book is your course using (assuming that this is homework)?
Assuming that your proof system does have modus tollens as a basic rule (not a usual choice, though!), then your proof sketch is fine, though it will need to be properly laid out according to the rules of the proof-system you are officially using. Otherwise you will need something more like this:

$ P \to Q\\ 
\quad\quad | \quad \neg Q\quad\quad\quad\quad\text{assumption}\\
\quad\quad | \quad \quad | \quad P\quad\quad\ \ \text{assumption}\\
\quad\quad | \quad \quad | \quad Q\\
\quad\quad | \quad \quad | \quad \text{contradiction!}\\
\quad\quad | \quad \neg P\\
 \neg Q \to \neg P$

A: Yes, it works, provided that you have Modus tollens available in your "rule-set".
For a proof of it which does not use modus tollens, according to the rules of Gordon J. Pace, Mathematics of Structures for Computer Science (2012), see page 45.
A: The OP asks:

Does anybody have a good reference to how I can learn how to proof such formulas (including equivalence, negation etc.)

You can use the Open Logic Project's proof checker online. They also have a textbook associated with the proof checker called forallx: Calgary Remix which is available to read online.

To derive $\lnot Q \to \lnot P$ from $P \to Q$ using modus tollens assume $\lnot Q$ on line 2, then derive $\lnot P$ with modus tollens (MT) on line 3  and finally derive the goal on line 4 with conditional introduction:

Without modus tollens one can derive this using Peter Smith's approach in his answer:

Again $\lnot Q$ is assumed on line 2 anticipating using conditional introduction at the end of the proof. Then make another assumption to use modus ponens (conditional elimination) on line 4. Now we have two lines that are contradictory. The absurdity symbol on line 5 allows us to discharge the assumption on line 3 and derive $\lnot P$ on line 5.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Fall 2019. http://forallx.openlogicproject.org/forallxyyc.pdf
