Understanding the Proof that Algebraic Integers are a Subring of $\Bbb{C}$

Because $f(\alpha)=0$, any polynomial $g(\alpha)$ reduces to a polynomial $r(\alpha)$ where $r$ has degree $<n$. Specifically, take $r$ to be the remainder when $g$ is divided by $f$ (in the ring of polynomials). So $g(x)=q(x)f(x)+r(x)$ and therefore $g(\alpha)=r(\alpha)$.

I personally like to also think of the ring $\mathbb{Z}\left[\alpha\right]$ as the smallest subring of the integers that contains $\alpha$. In this way, it's not so strange that $\mathbb{Z}\left[\alpha+\beta\right], \mathbb{Z} \left[\alpha \beta \right] \subseteq \mathbb{Z}\left[\alpha, \beta \right]$, since any polynomial in the sum or product of $\alpha$ and $\beta$ may definitely be written as a polynomial in the "mixed" monomials $\alpha^i \beta^j$. As for your second question, we could allow such monomials where $i+j\geqslant m+n$, but in this case at least one of $i$ and $j$ would be larger than $n$ or $m$, respectively. In the case where, say, $i \geqslant n$, we can then write $\alpha^i$ as a linear combination of lower order powers of $\alpha$, because $\alpha$ satisfies a polynomial equation of degree $n$. Thus we really don't need indices $(i,j)$ with their sum larger than $n+m-1$.