Biorthogonal Wavelet Expansion
Let me try to give you a short introduction to the concept of bi-orthogonal bases from a mathematical point of view. You are already using the concept, but your notation and nomenclature is obfuscating what is really going on. I will make this answer independent of the wavelet context you have, so that it is more general. You will certainly be able to apply it to your situation however.

You have a non-orthogonal basis, i.e. a set of linearly independent vectors spanning your $L^2(K)$ Hilbert space $\mathcal{H}$. That vector space induces another vector space, that of all linear forms $F:\mathcal{H}\to K$. We can make this space a dual space of $\mathcal{H}$ if we take the Hilbert space inner product $\langle \cdot, \cdot\rangle : \mathcal{H} \times \mathcal{H} \to K $ to generate a linear form $\langle \phi, \cdot \rangle : \mathcal{H} \to K$ for every $\phi \in \mathcal{H}$.

In an effort to simplify the notation a bit and make things more intuitive, we can introduce Dirac's Bra-Ket notation for vectors and their duals. Instead of $\phi \in \mathcal{H}$, we write the Ket $\left| \phi \right\rangle$ and instead of $\langle \phi , \cdot \rangle \; \phi \in \mathcal{H}$ we write the Bra $\left\langle \phi \right|$. Applying the linear form $\left\langle \psi\right|$ to $\left| \phi \right\rangle$ is then simply the notational fusion to a braket standing for the inner product $\left\langle \psi | \phi \right\rangle$.

In the bi-orthogonal basis business, this distinction is often ignored, making everything more confusing that it should be. Consider a basis that is not orthogonal with respect to the canonical inner product. We can of course use the usual knowledge of linear algebra to still construct everything of interest and expand vectors with respect to this basis, but it tends to get messy quickly. 

That's where a neat trick comes in. We simply invent a new inner product, i.e. a non-degenerate, non-negative bilinear form $B: \mathcal{H} \times \mathcal{H} \to  K$, so that $B( \phi_n, \phi_m ) = \delta_{n,m}$ for your basis $\{ \phi_n, n\in\mathbb{N}\}$ and the Kronecker symbol $\delta$. Because the basis is complete, $B$ is entirely determined by this condition. We can also write this bilinear form using a linear operator $B:\mathcal{H}\to\mathcal{H}$ with the same name. Then we get $$\left\langle \phi_n | B | \phi_m \right\rangle := \left\langle \phi_n | B \phi_m \right\rangle = B( \phi_n, \phi_m ) = \delta_{n,m}$$.

At this point, it makes sense to define the adjoint $A^\dagger$ of a linear operator $A$ like follows: $$\left\langle A^\dagger \phi_n | \phi_m \right\rangle = \left\langle \phi_n | A \phi_m \right\rangle$$

Of course it follows from the definitions above: $$\left\langle B^\dagger \phi_n \right| = B(\phi_n,\cdot)$$
and therefore $$\left\langle B^\dagger \phi_n | \phi_m \right\rangle=\delta_{n,m}$$
which means $\left\langle B^\dagger \phi_n \right|$ is the dual of $\left|\phi_n\right\rangle$ under $B$, whereas the dual under the canonical inner product is still $\left\langle \phi_n \right|$.

Now what you can often see, and I cannot stress enough how much I think this is a bad idea, is that the dual under $B$ is mapped back to $\mathcal{H}$ as a dual under the canonical inner product. That means you construct a second basis $\left| \tilde{\phi}_n\right\rangle =\left| B \phi_n \right\rangle$ which is bi-orthogonal to the first basis with respect to the canonical inner product: 
$$ \left\langle \tilde{\phi}_n | \phi_m \right\rangle = \delta_{n,m}$$
This obfuscates the real meaning of the second basis and puts both on the same footing and suggests that both bases live in the same space and transform in the same way, which they don't.

So, what is the alternative? Not taking the $B$-dual back to $\mathcal{H}$ makes things much more intuitive. You want to expand $\left|\psi\right\rangle$ in the basis $\phi_n$? Just use the $B$-dual: $\left|\psi\right\rangle=\sum_n \left| \phi_n \right\rangle \left\langle B^\dagger \phi_n | \psi \right\rangle $. Or you want to project onto a vector $\left| \kappa \right\rangle$? Just use the $B$-dual: $\left| \kappa \right\rangle \left\langle B^\dagger \kappa | \psi \right\rangle$. So whatever you do, just use $B$ as your inner product and you can work like with an orthogonal basis.

And to come back to your original question. You say that $V_0$ is not orthogonal to $\oplus_n W_n$. Yes, but only with respect to the canonical inner product. With respect to $B$, they are orthogonal. And so your argument for the orthogonal basis case works.

I hope all this does not confuse you. I've tried to be as clear as possible, but it's not a trivial topic, so don't be frustrated if you have to read the answer several times until it makes sense. If I can clarify anything, please let me know.