I think I understand this much: Let $X$ be a Banach space and $L(X)$ the space of bounded linear operators on $X$. Let $\overline{X} = \{x \in X : \|x\| = 1\}$. For operators $T, S \in L(X)$ let $\sup_{x \in \overline{X}} \|Tx - Sx \| = \varepsilon$. Then the magnitude of $\varepsilon$ gives us some intuition about how 'close together' operators $T$ and $S$ are in $L(X)$ with the norm topology. (Although if that much is wrong, please let me know!)
But is there a similar intuition for 'closeness' when $L(X)$ is given the strong operator topology?
So the way to think about it is this:
if you have a sequence $T_n \to T$ of operators in the norm topology, then this means that, as you have noted, $$\sup_{|x|=1} |\!|T_nx-Tx|\!| \to 0,$$ that is to say the convergence is uniform.
The strong operator topology is defined to be the smallest topology that makes the evaluations $ev_x$ that send $A \mapsto A(x)$ continuous. If you unwind a little, this means that $T_n\to T$ is the same thing as $$|\!|T_nx-Tx|\!| \to 0 \quad \text{for all } x.$$ So in this sense, closeness in the strong topology means that you are pointwise close, while the norm topology tells you if your operators are uniformly close.