This thread is just a note.
Given Hilbert spaces.
Then boundedness will be inherited: $$A,B\text{ bounded}\implies A\otimes B\text{ bounded}$$ Especially, the bounds multiply: $$\|A\otimes B\|=\|A\|\cdot\|B\|$$
Clearly for simple tensors: $$\|(A\otimes B)(\varphi\otimes\psi)\|=\|A\varphi\|\cdot\|B\psi\|\leq\|A\|\cdot\|B\|\cdot\|\varphi\|\cdot\|\psi\|=\|A\|\cdot\|B\|\cdot\|\varphi\otimes\psi\|$$ But what about arbitrary tensors?
This answer is community wiki.
Take an orthonormal basis: $$\mathcal{S}\otimes\mathcal{T}:=\{\sigma\otimes\tau:\sigma\in\mathcal{S},\tau\in\mathcal{T}\}$$
Then one has the bound: $$\left\|A\otimes B\left(\sum_{\sigma\tau}\lambda_{\sigma\tau}\sigma\otimes\tau\right)\right\|=\left\|\sum_{\sigma\tau}\lambda_{\sigma\tau}A\sigma\otimes B\tau\right\|\leq\sum_{\sigma\tau}|\lambda_{\sigma\tau}|^2\|A\sigma\otimes B\tau\|\\=\sum_{\sigma\tau}|\lambda_{\sigma\tau}|^2\|A\sigma\|\| B\tau\|\leq\|A\|\|B\|\sum_{\sigma\tau}|\lambda_{\sigma\tau}|^2\|\sigma\|\|\tau\| =\|A\|\cdot\left\|\sum_{\sigma\tau}\lambda_{\sigma\tau}\sigma\otimes\tau\right\|$$ (In fact, this should be rather taken carefully on finite sums before.)