Grothendieck group "commutes" with direct sum

Here's a sketch. You have to construct the inverse using universal properties as well. You have a composition $$M \hookrightarrow M\oplus N \stackrel{i_{M\oplus N}}{\longrightarrow} \mathcal{G}(M\oplus N)$$which induces a map $\mathcal{G}(M) \to \mathcal{G}(M\oplus N)$. Similarly, you get a map $\mathcal{G}(N) \to \mathcal{G}(M\oplus N)$. The universal property of the direct sum joins these two maps into a map $\psi\colon\mathcal{G}(M)\oplus \mathcal{G}(N) \to \mathcal{G}(M\oplus N)$. Now you have two maps $$\psi\circ \varphi\colon \mathcal{G}(M\oplus N) \to \mathcal{G}(M\oplus N)\quad\mbox{and}\quad\varphi\circ \psi\colon \mathcal{G}(M)\oplus \mathcal{G}(N) \to \mathcal{G}(M)\oplus \mathcal{G}(N).$$Using the uniqueness provided by the universal properties of $\mathcal{G}$ and $\oplus$, argue that these compositions equal the identity. It works the same for defining maps $$\bigoplus_{\alpha} \mathcal{G}(M_\alpha) \to \mathcal{G}\left(\bigoplus_{\alpha}M_\alpha\right) \quad\mbox{and}\quad \mathcal{G}\left(\bigoplus_\alpha M_\alpha\right) \to \bigoplus_\alpha \mathcal{G}(M_\alpha)$$and running the above argument through.

The Grothendieck completion is the left adjoint of the forgetful functor from Abelian groups to commutative monoids. That is, if $\mathcal{M}$ denotes the functor $\mathcal{M}\colon\mathfrak{A}\to\mathfrak{M}$ from abelian groups to commutative monoids that maps the abelian group $G$ to itself considered as a monoid, then for any commutative monoid $M$ and abelian group $G$, we have a natural isomorphism $$\mathfrak{M}(M,\mathcal{M}(A))\cong \mathfrak{A}(\mathcal{G}(M),A).$$

Left adjoints respect colimits, right adjoints respect limits. As the direct sum is a coproduct/colimit, it follows that $\mathcal{G}$ respects direct sums. Explicitly, recall that a map from a direct sum is equivalent to maps from each constituent: each morphism $f\colon\oplus_{\alpha\in A}X_{\alpha}\to Z$ corresponds to a family of maps $\{ f_{\alpha}\colon X_{\alpha}\to Z\}_{\alpha\in A}$ (in any category in which the direct sum is a coproduct; if not, then you should use the coproduct instead of the direct sum). Thus, for every abelian group $A$, $$\begin{align*} \mathfrak{A}(\mathcal{G}(\oplus_{\alpha}M_{\alpha}),A) &\cong \mathfrak{M}(\oplus_{\alpha}M_{\alpha},\mathcal{M}(A))\\ &\cong \prod_{\alpha}\mathfrak{M}(M_{\alpha},\mathcal{M}(A))\\ &\cong \prod_{\alpha}\mathfrak{A}(\mathcal{G}(M_{\alpha}),A)\\ &\cong \mathfrak{A}(\oplus_{\alpha}\mathcal{G}(M_{\alpha}),A). \end{align*}$$ This means that $\mathcal{G}(\oplus_{\alpha} M_{\alpha})$ has the universal property of $\oplus_{\alpha}\mathcal{G}(M_{\alpha})$, hence the two are isomorphic.

Cf. the proof that the free monoid on a disjoint union of two sets is the coproduct of the free monoids on each set. It’s the same, because the relation between the “free monoid” construction and the “underlying set” functor is the same as the relation between the “Grothendieck completion group” construction and the “underlying monoid” functor. This in turn is a special case of the result mentioned above, that left adjoints respect colimits and right adjoints respect limits. Which is one reason why you should care about adjoint functors.