Let $A,B \in \mathbb{R}^{m \times n}$ and let $\|A\|_2=\sqrt{\lambda_{max}(A^\mathsf{T}A)}$ denote the spectral norm of a matrix, where $\lambda_{max}$ is the maximum eigenvalue.
Clearly, we have $\|A+B\|_2\leq \|A\|_2+\|B\|_2$ by the triangle inequality and also $$\|\begin{bmatrix}A&B\end{bmatrix}\|_2 \leq \|A\|_2+\|B\|_2.$$
Therefore, $\|A+B\|_2$ and $\|\begin{bmatrix}A&B\end{bmatrix}\|_2$ share the same upperbound. We can conclude that there is a large enough constant $K>0$ such that $$\|A+B\|_2\leq K~~\|\begin{bmatrix}A&B\end{bmatrix}\|_2\leq K\|A\|_2+K\|B\|_2\,.$$
However, this relationship alone is not very useful, as we don't know anything about $K$. Is it possible to characterize $K$ based on $A$ and $B$, or is there another useful bound linking $\|A+B\|_2$ and $\|\begin{bmatrix}A&B\end{bmatrix}\|_2$?
$$ \|A+B\|_2 =\left\|\pmatrix{A&B}\pmatrix{I\\ I}\right\|_2 \le\|\pmatrix{A&B}\|_2\left\|\pmatrix{I\\ I}\right\|_2 =\sqrt{2} \|\pmatrix{A&B}\|_2. $$ This bound is sharp as it is attained by $A=B=I$ when $m=n$, $A=B=\pmatrix{I&0}$ when $m<n$ or $A=B=\pmatrix{I\\ 0}$ when $m>n$.