$M$ and $N$ equal in $K_0$ $\Rightarrow$ $\exists P$ such that $M\oplus P\cong N\oplus P$ Let $\mathcal P(A)$ be the category of finitely generated projective $A$-modules ($A$ is a ring with unity). Then consider the free group $F$ over the isomorphism classes of $\mathcal P(A)$. I will indicate the isomorphism classes with  square brackets $[\cdot]$.
It means that an element of $F$ can be written as a finite sum:
$$\sum_k n_k[M_k]$$
where $n_k\in\mathbb Z$.
$H$ is the subgroup of $F$ generated by the elements of the type:
$$[M\oplus N]-[M]-[N]$$

Now please help me to understand why the following implication is true:

$[M]=[N] \operatorname{mod } H\Rightarrow$ there exists $P\in\mathcal P(A)$ such that $[M\oplus P]=[N\oplus P]$ 


By the definition of a free abelian group, we have the following observation:
If $[M_1] + \dotsb + [M_r] = [N_1] + \dotsb + [N_s]$, we have $r=s$ and the summands are the same up to permutations. Since the direct sum is commutative, we get an isomorphism.
$$M_1 \oplus \dotsb \oplus M_r \cong N_1 \oplus \dotsb \oplus N_s.$$
Using this, the proof is straight forward: By definition you find $A_i, B_i$ and $C_j, D_j$, such that
$$[M]-[N] = \sum_{i=1}^m \Big([A_i \oplus B_i]-[A_i]-[B_i]\Big) - \sum_{i=1}^n \Big([C_j \oplus D_j]-[C_j]-[D_j]\Big),$$
or equivalently:
$$[M] + \sum_{i=1}^m \Big([A_i]+[B_i]\Big) + \sum_{j=1}^n [C_j \oplus D_j] = [N] + \sum_{i=1}^m [A_i \oplus B_i] + \sum_{i=1}^n \Big([C_j]+[D_j]\Big)$$
Using the observation discussed above, we get the desired result with
$$P = \bigoplus_{i=1}^m \Big(A_i \oplus B_i\Big) \oplus \bigoplus_{j=1}^n \Big(C_j \oplus D_j\Big).$$
