Suppose $\alpha$ is a root of polynomial $P(x)$ of degree $n$, and $\beta$ is a root of polynomial $Q(y)$ of degree $m$ with coefficients in $\mathbb{Q}$. Using $P(x)$ every power of $\alpha$ can be reduced to a linear combination of $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$, similarly every power of $\beta$, modulo $Q(y)$ is a linear combination of $\{1,\beta, \ldots, \beta^{m-1}\}$. Coefficients in these linear combinations are rational numbers.
Consider $\omega = \alpha + \beta$. We can constructively prove that this number is algebraic. Using reductions of powers of $\alpha$ and $\beta$ we can represent any power of $\omega$ as $\sum_{p=0}^{n-1} \sum_{q=0}^{m-1} c_{p,q}\alpha^p \beta^q$, with $c_{p,q} \in \mathbb{Q}$. Thus powes of $\omega$ are mapped into $\mathbb{Q}^{m \cdot n}$, and therefore there will exist coefficients $\{a_0, a_1, \ldots, a_{m \cdot n}\} \in \mathbb{Q}$ such that
$$
a_0 +a_1 \omega + \cdots + \omega^{m \cdot n} a_{m \cdot n} = 0
$$
Example: For $\phi$, polynomial is $P(T) = T^2-T-1$ and for $\sqrt{2}$ the polynomial is $Q(T) = T^2 - 2$. Thus, using basis $\{1, \phi, \sqrt{2}, \phi \sqrt{2}\}$:
$$
\begin{eqnarray}
1 &=& 1 \cdot 1 + 0 \cdot \phi + 0 \cdot \sqrt{2} + 0 \cdot \phi \sqrt{2} \\
\phi+\sqrt{2} &=& 0 \cdot 1 + 1 \cdot \phi + 1 \cdot \sqrt{2} + 0 \cdot \phi \sqrt{2} \\
\left(\phi+\sqrt{2}\right)^2 &=& 3 \cdot 1 + 1 \cdot \phi + 0 \cdot \sqrt{2} + 2 \cdot \phi \sqrt{2} \\
\left(\phi+\sqrt{2}\right)^3 &=& 1 \cdot 1 + 8 \cdot \phi + 5 \cdot \sqrt{2} + 3 \cdot \phi \sqrt{2} \\
\left(\phi+\sqrt{2}\right)^4 &=& 18 \cdot 1 + 15 \cdot \phi + 4 \cdot \sqrt{2} + 16 \cdot \phi \sqrt{2}
\end{eqnarray}
$$
Solving the null space problem, we find that $\phi + \sqrt{2}$ is a zero of
$$
z^4 -2 z^3 - 5 z^2 + 6z -1 = 0
$$
Similarly, one can establish the polynomial whose root is $\phi \cdot \sqrt{2}$, which is $z^4 - 6 z^2 + 4$. In Mathematica, one can use MinimalPolynomial (online ref-page) to do these computations.
