Motivation for norm of linear operator $T:E \rightarrow F$ defined for $\|x\|_E \leq 1$?
Initial guess: standardization/normalization. I.e. this gives the "unit measure" of the norm.
See def. here:
Showing that $\|T\| = \sup\{\|Tx\| : \|x\| \leq 1 \} = \sup\{\|Tx\| : \|x\| = 1 \}$
On
The reason why we don't define $$ \Vert T \Vert_\infty = \sup_{x\in E} \Vert Tx \Vert_F $$ is that this quantity is either zero or $+\infty$. Indeed, if $T$ is not equal to the zero function, then there exists $x_0$ such that $Tx_0 \neq 0$. However, then we get $$ \Vert T \Vert_\infty \geq \sup_{t\in \mathbb{R}} \Vert T(tx_0)\Vert_F = \sup_{t\in \mathbb{R}} \vert t \vert \cdot \Vert Tx_0 \Vert_F = + \infty. $$ Of course we could define $$ \Vert T \Vert_C = \sup_{x\in E: \ \Vert x \Vert_E \leq C} \Vert Tx \Vert_F $$ for your favourite $C>0$ but this is essentially the same thing as the usual definition of the norm as we have $$\Vert T \Vert_C = \sup_{x\in E: \ \Vert x \Vert_E \leq C} \Vert Tx \Vert_F = \sup_{x\in E: \ \Vert x \Vert_E \leq 1} \Vert T(Cx) \Vert_F = C \Vert T \Vert $$
Suppose that there is $C>0$ s.t. $$\|Tx\|_{F}\leq C\|x\|_E.$$ In particular, if $\|x\|_E\neq 0$, then $$\frac{\|Tx\|_{F}}{\|x\|_E}\leq C,$$ i.e. $$\sup_{x\neq 0}\frac{\|Tx\|_F}{\|x\|_E}\leq C<\infty .$$
We therefore define $\|T\|$ by the smallest $C$ s.t. $$\|Tx\|_F\leq C\|x\|_E,$$ for all $x\in E$, or equivalently as $$\|T\|:=\sup_{x\neq 0}\frac{\|Tx\|_F}{\|x\|_E}.$$
I let you show that it's indeed a norm. In fact, since $T$ is linear, $$\frac{1}{\|x\|_E}T(x)=T\left(\frac{x}{\|x\|_E}\right),$$ and thus $$\frac{\|Tx\|_F}{\|x\|_E}= \left\|T\left(\frac{x}{\|x\|}\right)\right\|_F=\|T(y)\|_F,$$ where $\|y\|_E=1$. Therefore, we have that $$\|T\|=\sup_{\|y\|_E=1}\|T(y)\|_F,$$
and thus, $$\|T\|\leq \sup_{\|x\|_E\leq 1}\|Tx\|_{F}.$$
I let you prove (as an exercise) that the inverse inequality holds as well.