How can two orbitals constructively and destructively interfere simultaneously?
They do not simultaneously interfere constructively and destructively.

Let me explain. Since the time-independent Schrödinger equation is not separable for molecules due to inter-electronic interaction terms, we can seek solutions to this problem with the variational method.

Let's examine the case of the $\mathrm{H_2}$ molecule. The most natural/physically meaningful/simple choice is to use a trial function that consists of a linear combination of hydrogen 1s atomic orbitals:
$$E\left(c_{1},c_{2}\right)=\dfrac{\left\langle c_{1}\psi_{1}+c_{2}\psi_{2}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|c_{1}\psi_{1}+c_{2}\psi_{2}\right\rangle }{\left\langle c_{1}\psi_{1}+c_{2}\psi_{2}|c_{1}\psi_{1}+c_{2}\psi_{2}\right\rangle }=\dfrac{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}\left\langle \psi_{i}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\psi_{j}\right\rangle }{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}\left\langle \psi_{i}|\psi_{j}\right\rangle }$$

$$H_{ij}\equiv\left\langle \psi_{i}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\psi_{j}\right\rangle$$

$$S_{ij}\equiv\left\langle \psi_{i}|\psi_{j}\right\rangle$$

$$E\left(c_{1},c_{2}\right)=\dfrac{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}H_{ij}}{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}S_{ij}}$$

After minimizing the energy with respect to $c_1$ and $c_2$, you will end up with two energy levels, each one associated with a wave function:
$$E_{+}=\dfrac{H_{11}-H_{21}}{\left(1-S_{12}\right)}\qquad\psi_{+}=\dfrac{1}{\sqrt{2}}\left(\psi_{1}+\psi_{2}\right)$$

$$E_{-}=\dfrac{H_{11}+H_{21}}{\left(1+S_{12}\right)}\qquad\psi_{-}=\dfrac{1}{\sqrt{2}}\left(\psi_{1}-\psi_{2}\right)$$

In other words, by choosing to model the $\mathrm{H_2}$ molecule with two hydrogen $1s$ atomic orbitals, you find two discrete energy levels, each one associated with its wave function (one anti-bonding and one bonding). So this answers your questions:


The two atomic orbitals do not simultaneously interfere: The MOs are two different wave functions.
Of course the anti-bonding orbital exists when not occupied.


Note that the previous procedure can be generalized for all kind of atomic orbitals.

I want to clarify something about the meaning of the wave function since you're asking: As a result, is it correct to think that it's the orbitals themselves that are the waves, not the electrons?

Let's assume that your system consists of a single particle in a one-dimensional space. The wave function, $\Psi\left(x,t\right)$,
is a complex function that contains all the information about the system. But what is really physical meaningful is its modulus squared, $\left|\Psi\left(x,t\right)\right|^{2}$, which represents the position probability density. What is the position probability density? It's something that multiplied by a certain length (or volume if you're working in three dimensions), let's say dx, gives you the probability for finding the particle between $x$ and $x+dx$
$$P\left(x,x+dx\right)=\left|\Psi\left(x,t\right)\right|^{2}dx$$.
Of course, you can integrate the previous equation to find the probability over a certain region of space, $\Omega$
$$P_{\Omega}={\displaystyle \int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}dx$$.