I have some naive questions (I'm a beginner): I consider the space $L^2$ of functions $\mathbb R\to\mathbb R$ equippped with the usual inner product $\langle \cdot,\cdot\rangle$. Let $V$ be a subspace of $L^2$ and let $\{\phi_k\}_{k\in\mathbb Z}$ be a Riesz basis for $V$. Then there exists a Riesz basis $\{\tilde\phi_k\}_k$ dual to $\{\phi_k\}_k$. By definition there exists an orthonormal basis $\{e_k\}_k$ for $V$ and a bounded bijective operator $U:L^2\to L^2$ such that $\phi_k=Ue_k$ and $\tilde \phi_k = (U^{-1})^*e_k$, having denoted with $(U^{-1})^*$ the adjoint of $U^{-1}$.
First question: By definition I have \begin{align*} \delta_{k,n} = \langle e_k,e_l\rangle = \langle U^{-1}\phi_k,U^{-1}\phi_n\rangle=\langle\phi_k,(U^{-1})^*U^{-1}\phi_n\rangle \end{align*} Hence, with $C:=(U^{-1})^*U^{-1}$, and with $\langle\cdot,\cdot\rangle_C$ the inner product $\langle f,g\rangle_C := \langle f, Cg\rangle$, I have $$ \langle \phi_k,\phi_n\rangle_C=\delta_{k,n} $$ Can I say that $\{\phi_k\}_k$ is an orthonormal with respect to the inner product $\langle\cdot,\cdot\rangle_C$ basis for $V$?
Now, let $W$ be another subspace of $L^2$ and let $\{\psi_k\}_k$ be a Riesz basis for $W$. Assume that, for each $k,n\in\mathbb Z$, $\langle\psi_k,\tilde\phi_n\rangle=0$.
Second question: By definition I have: \begin{align*} \langle \psi_k,\phi_n\rangle_C &= \langle \psi_k, C\phi_n\rangle = \langle \psi_k, (U^{-1})^* U^{-1}\phi_n\rangle\\&= \langle\psi_k, (U^{-1})^* e_n\rangle = \langle\psi_k,\tilde\phi_n\rangle = 0 \end{align*} Is it appropriate to say that $V\perp W$ in the inner product $\langle\cdot,\cdot\rangle_C$?
Yes, a Riesz basis becomes orthonormal if the inner product is changed to fit the basis, as you did here.
Note that the existence of $U$ is not exactly "by definition". The definition of Riesz basis provides an invertible operator $V\to V$, which then needs to be extended to $L^2$ in some invertible way. linear obtained by extending a bijective.
The second conclusion is also correct. It becomes more transparent if one notices that $C$ maps $V$ onto itself, so if a vector is orthogonal to $V$ in the usual sense, it is also orthogonal to it in the sense of the $C$-inner product.