Conjugate transpose of matrix is the adjoint intuition I'm having a bit of trouble understanding this fact from Linear Algebra Done Right

Let T $\in \mathcal{L}(V, W)$ Suppose $e_{1}, \dots, e_{n}$ is an
  orthonormal basis of $V$ and $f_{1}, \ldots, f_{m}$ is an orthonormal
  basis of $W$. Then then adjoint $T^{*}$ is the conjugate transpose of
  $T$ in $\langle T v, w\rangle=\left\langle v, T^{*} w\right\rangle$

I went through the proof and did an example but I can't get a clearer picture as to why this must be true intuitively or geometrically. What is so special about a transpose that makes the transformation act in this bridge-like fashion?
Also, I am also not able to appreciate the fact that we have an orthonormal basis, while it is important in the proof, what would go wrong if we didnt? 
Here's the entire proof for reference:

Any inner product $\langle v, w\rangle$ on a real vector space has a representation as a symmetric positive-definite matrix $M$, with
$$\langle v, w\rangle = v^T M w.$$
Now if $T$ and $T^*$ are adjoint you must have
$$\langle Tv, w\rangle = v^T T^TMw = v^T MT^* w = \langle v, T^*w\rangle.$$
Since $v$ and $w$ are arbitrary, it follows that
$$T^TM = MT^*.$$
In the special case of the Euclidean inner product, $M=I$ and the above reduces to $T^T=T^*$. Notice that this relationship is not true for general inner products $M$.
Now if you have two different inner product spaces $V$, $W$ with inner products
$$\langle v_1, v_2\rangle_{M_v}, \langle w_1, w_2\rangle_{M_w}$$
and $T: V\to W$, an identical calculation shows that $T$ and $T^*$ satisfy
$$T^TM_{w} = M_v T^*$$
with $T^T = T^*$ in the special case that both inner products are Euclidean.
