How does the long exact homology sequence coming from the mapping cone look like? 
3.3.1. Let $f \colon P_\bullet \to Q_\bullet$ be a chain map of complexes.
  We define the mapping cone in the following way.
  Let $M_n = P_{n-1} \oplus Q_n$, and define $d_n^M \colon M_n \to M_{n-1}$ by
  $$
  d_n^M(x,y) = ( -d_{n-1}^P(x), d_n^Q(y) + f(x) ).
$$
  Show that $(M_n, d_n^M)$ forms a chain complex, and that we have a long exact sequence
  $$
  \dotsb
  \longrightarrow
  \operatorname{H}_n(Q)
  \longrightarrow
  \operatorname{H}_n(M)
  \longrightarrow
  \operatorname{H}_{n-1}(P)
  \longrightarrow
  \dotsb
$$
(Original picture of the problem here.)

This is an exercise so I am not looking for any answers but rather to understand parts of the question I am uncertain of. I have managed to prove that we in fact have a complex, however I don’t quite understand how the arrows in the long exact sequence looks like, can someone bring clarity to this?
 A: I will write $Cf$ for the cone of $f$ (which is your complex $M$, not a morphism). Note that from $f : P\to Q$ we obtain a mapping $Hf :HP \to HQ$ which assigns a class $[x]$ to $[fx]$. Moreover, we have a short exact sequence of complexes of the form $0\to Q\to C(f)\to P\to 0$ where the nontrivial maps are the inclusion and the projection. The long exact sequence is now obtained from the usual construction using the snake lemma, so in particular the arrows $HQ\to HCf$ and $HCf\to HP$ are induced by the inclusion and the projection. Part of the point of all this is that the connecting morphism of the SEC above is precisely $Hf$.
To see this, let's follow the recipe of the snake lemma: suppose that we have $x\in P$ such that $dx=0$, and lift this to $(x,0)$ in $Cf$. If we apply the differential of $Cf$ we obtain $(0,fx)$, and we can lift this to $fx$ in $Q$. Taking classes shows that $\delta[x] = [fx]$, as desired. 
A corollary of all this is that $Hf$ is a quasi-isomorphism if and only if $Cf$ is acyclic. You can also show that $f$ is an homotopy equivalence if and only if $Cf$ is contractible, too. So $Cf$ plays the same role as the usual mapping cone of topological spaces. 
