Suppose that $H$ is a Hilbert space and $W\subset H$ a closed subspace. Let $T\colon H\to H$ be an operator. Suppose that $T_{W}\colon W\to H$ and $T_{W^{\perp}}\colon W^{\perp}\to H$ are bounded. Given this data, can we explicitly compute the operator norm $\|T\|$ of $T$ (in terms of the operator norms of $T|_{W}$ and $T|_{W^{\perp}}$)?
Since $W$ is a closed subspace, I think one has to use the fact that $H$ is isomorphic (as Hilbert spaces) to the (algebraic) direct sum $W\oplus W^{\perp}$, where $W\oplus W^{\perp}$ is endowed with the innerproduct $$\langle u\oplus u^{\perp},v\oplus v^{\perp}\rangle_{W\oplus W^{\perp}}:=\langle u,v\rangle_{H}+\langle u^{\perp},v^{\perp}\rangle_{H},$$ where $u,v\in W$ and $u^{\perp},v^{\perp}\in W^{\perp}$.
No. Even in two dimensions this would not be enough information. Consider the transformations given by matrices $\begin{bmatrix} 1&0\\0&1\end{bmatrix}$ and $\begin{bmatrix} 1&1\\0&0\end{bmatrix}$, where $W$ and $W^\perp$ are the coordinate axes. In both cases both restricted norms are $1$, but the norm is $1$ in one case and $\sqrt2$ in the other.
You can get lower and upper bounds as follows. Obviously $\|T\| \geq \max\{\|T|_W\|,\|T|_{W^\perp}\|\}$. For the upper bound, consider a unit vector $v \in H$ and write $v=w+x$ where $w\in W$ and $x \in W^{\perp}$.
$$ \|Tv\| = \|Tw+Tx\| \leq \|Tw\|+\|Tx\| \leq \|T|_W\|\|w\|+ \|T|_{W^\perp}\|\|x\|$$
The right side is maximized when the ratio between $w$ and $x$ is the same as the ratio between $\|T|_{W}\|$ and $\|T|_{W^\perp}\|$. So
$$ \|T\| \leq \frac{\|T|_{W}\|^2 + \|T|_{W^\perp}\|^2}{\sqrt{\|T|_{W}\|^2 + \|T|_{W^\perp}\|^2}} = \sqrt{\|T|_{W}\|^2 + \|T|_{W^\perp}\|^2}$$