Definition of the differential of the Cone of a morphism of complexes Let $(F^\bullet,d_F)$ and $(G^\bullet,d_G)$ be two complexes in an abelian category $\mathbf{A}$.
The complex cone $Cone(\varphi)^\bullet$ of a morphism of complexes $\varphi:F^\bullet \to G^\bullet$ is defined as
$$Cone(\varphi)^i=G^i\oplus F^{i+1},$$
and its differential is
$$d(g^i,f^{i+1})=(d_G(g^i)+\varphi(f^{i+1}),-d_F(f^{i+1})).$$
then there are natural maps $G^\bullet \to Cone(\varphi)^\bullet$ and $Cone(\varphi)^\bullet \to F[1]^\bullet$ that make
$$F\to G\to Cone(\varphi) \to F[1]$$
into a distinguished triangle inside the derived category $\mathbf{D}^b(\mathbf{A})$.
My question is: what is the reason behind the "twisting" of the first component of the differential with $\varphi(f^{i+1})$? Shouldn't one obtain an honest complex even without that? It must be required by some interesting property of the cone itself.
 A: Short answer: Otherwise it wouldn't depend on $\phi$!
Longer answer: Think about it this way: write $F$ and $G$ vertically side by side (in the 0-th and 1st column, respectively), with horizontal maps $\phi$. Since $\phi$ commutes with $d$, you get a double complex, call it $C$. The projection to $F$ and the inclusion of $G$ give a short exact sequence of double complexes $$0
\to G[-1] \to C\to F[0]\to 0.$$ Here $G[-1]$ means $G$ considered as a double complex in the 1st column, and similarly for $F[0]$. Now we take the total complexes. Recall that for a double complex $(C^{p,q}, d^v, d^h)$, ${\rm Tot}(C^{\bullet, \bullet})$ is the complex $K^n = \oplus_{p+q=n} C^{p,q}$, $d=d^h + (-1)^p d^v$. The sign twist is to make the squares in $C^{p, q}$ anticommute, so that $d^2=0$ in the total complex. After this operation, we get a short exact sequence of complexes $$ 0\to G[-1] \to K\to F\to 0 $$ where $K={\rm Tot}(C)$, and now $G[-1]$ means $G$ shifted by 1 to the right. Now you can check that 


*

*$K= {\rm Cone}(\phi)[-1]$, 

*the boundary maps $\delta:H^i(F)\to H^{i+1}(G[-1])=H^i(G)$ in the long cohomology exact sequence  equal $H^i(\phi)$. 


This explains why we have an exact triangle as desired.
I apologize for any potential sign errors.
A: You might want to take a look at Problem 1 on the following problem set from a course on homological algebra.  http://www.math.stonybrook.edu/~jstarr/M536f15/M536f15ps10.pdf 
In particular, the mapping complex satisfies a property (up to homotopy) that makes it seem like a kernel, and it simultaneously satisfies a property (up to homotopy) that makes it seem like a cokernel.
