If $H$ is finite dimensional this can be proved e.g. from the fact that commuting matrices over $\mathbb{C}$ must be simultaneously upper triangularizable, see e.g. this MSE question and answer, and the observation that the eigenvalues of an upper triangular matrix are its diagonal entries.
If $H$ is infinite dimensional (or even if $H$ is finite dimensional), one approach would use the fact that the spectra of $A$ and $B$ in $\mathcal{B}(H)$ are the same as the spectra of $A$ and $B$ relative to any maximal commutative Banach subalgebra $\mathcal{M}$ of $\mathcal{B}(H)$ that contains both $A$ and $B$, and the spectrum of an element $x$ of a commutative Banach algebra is the range of its Gelfand transform (i.e., the set $\{\phi(x): \text{$\phi$ is a nonzero multiplicative linear functional}\}$). The relevant generalities are already covered in this MSE answer (see in particular Theorem 2.6).
In slightly more detail, with that background, the calculation is then (fixing a maximal commutative $\mathcal{M}$ as above, and writing $\sigma_{\mathcal{A}}(x)$ for the spectrum of $x$ in $A$)
\begin{align*}
\sigma_{\mathcal{B}(H)}(A+B) & = \sigma_{\mathcal{M}}(A+B) \\
& = \{\phi(A+B): \text{$\phi: \mathcal{M} \to \mathbb{C}$ is a nonzero multiplicative linear functional} \} \\
& = \{\phi(A) + \phi(B): \text{$\phi: \mathcal{M} \to \mathbb{C}$ is a nonzero multiplicative linear functional} \}\\
& \subseteq \sigma_{\mathcal{M}}(A) + \sigma_{\mathcal{M}}(B) \\
& = \sigma_{\mathcal{B}(H)}(A) + \sigma_{\mathcal{B}(H)}(B) \\
\end{align*}
The first equality uses maximality of $\mathcal{M}$ (i.e. $\mathcal{M}$ contains anything in $\mathcal{B}(H)$ that commutes with both $A$ and $B$ in $\mathcal{B}(H)$, including any inverse of any element $(A + B) - \lambda I$ that is invertible in $\mathcal{B}(H)$), the second equality is the Theorem 2.6 mentioned above, the third equality is because each $\phi$ is linear, the subsequent inclusion is a use of Theorem 2.6 to conclude that $\phi(A) \in \sigma_{\mathcal{M}}(A)$ and $\phi(B) \in \sigma_{\mathcal{M}}(B)$ for any $\phi$, and the last equality is again by maximality of $\mathcal{M}$.
I do not have Kadison and Ringrose handy, so I cannot confirm if this is the type of argument that the authors of the paper that you cite had in mind for the general case, but it wouldn't surprise me if it was.
