Consider a basis of $H$, say $\{e_i\}$. Then, note that for any $x = \sum_{i=1}^n x_ie_i$ (where $n = \dim H$) we have $$||Tx||^2 = \langle Tx,Tx\rangle = \left\langle \sum_{i=1}^n x_i T(e_i),\sum_{i=1}^n x_i T(e_i) \right\rangle = \sum_{i,j=1}^nx_i \overline{x_j}\left\langle T(e_i),T(e_j)\right\rangle$$
Now, let $M = \displaystyle\max_{1 \leq i,j \leq n} \langle T(e_i),T(e_j)\rangle$, existing because we are taking the maximum over a finite set. Then, we have:
$$
||Tx||^2 = \sum_{i,j=1}^n x_i \overline{x_j}\langle T(e_i)T(e_j)\rangle \leq M \sum_{i,j=1}^n x_i \overline{x_j} \leq M ||x||^2
$$
taking square roots on both sides tells us that the norm of $T$ exists, and is less than or equal to $\sqrt M$.
This is not a geometric intuition. Nevertheless I find it pretty enlightning, so I hope to be helpful.
Given a linear operator $T:V\to V$, you can define the transpose operator:
$$T^t:V^*\to V^*$$
$$\ \ \ \ \ \ \ \ \ \varphi \mapsto \varphi\circ T$$
The transpose operator is essentially the simplest way you can imagine to construct an operator on $V^*$ using an operator $T$ on $V$.
If you understand this then the adjoint is straight forward!
In fact the idea is to use the fact that $V$ is isomorphic to its dual $V^*$. Normally there aren't natural isomorphism between a space and its dual, but in this case you are in a inner product (finite dimensional ) space!
So you can construct a canonical isomorphism(actually if the field is $\mathbb{C}$, this is conjugate-linear):
$$\omega:V\to V^*$$
$$\ \ \ \ \ \ \ \ \ v \mapsto \langle \bullet,v\rangle=\omega_v$$
(Basically you associate to a vector $v$ the linear functional $w\mapsto \langle w,v \rangle$).
Now the adjoint is simply the transpose "re-interpreted" through this natural isomorphism:
$$T^*=\omega^{-1}\circ T^t \circ \omega:V\to V$$
As you can see the adjoint is simply a natural reinterpretation of the transpose, so if you find the transpose intuitive, you should also find the adjoint intuitive.
Clearly I should prove that my definition of $T^*$ coincides with yours, and this is straightforward:
$$\langle T(x),y \rangle=\langle x,T^*(y) \rangle \ \text{for all } x,y \iff $$
$$\iff \omega_y(T(x))=\omega_{T^*(y)}(x) \ \text{for all } x,y \iff $$
$$\iff \omega_y\circ T=\omega_{T^*(y)} \ \text{for all } y \iff $$
$$\iff T^t(\omega_y)=\omega_{T^*(y)} \ \text{for all } y \iff $$
$$\iff (T^t\circ \omega)(y)=(\omega \circ T^*)(y) \ \text{for all } y \iff $$
$$\iff T^t\circ \omega=\omega \circ T^* \iff $$
$$\iff T^*=\omega^{-1}\circ T^t \circ \omega $$
Best Answer
Even if $W$ is not finite-dimensional, the range of $T$ will be a finite-dimensional subspace of $W$ (since it is spanned by the image of a basis for $V$), and that subspace is all that matters for the boundedness of $T$.
So without loss of generality we can assume $W$ is finite dimensional.
Choose orthonormal bases for $V$ and $W$, and write down the matrix for $T$. Then $\|T\|$ cannot possibly exceed the sum of the absolute values of the matrix entries.