How do atomic orbitals interfere both constructively AND destructively to give two molecular orbitals?
Part of the issue is that you're treating orbitals as objects. They aren't objects, they are quantum mechanical wavefunctions. The electrons which occupy the orbitals are objects, but the orbitals themselves are decidedly not objects.
When we talk about quantum mechanical states, the overall phase of an individual state has no meaning. It is a fundamental postulate of quantum mechanics that the wavefunction $\psi(x)$ and the wavefunction $k\cdot\psi(x)$ represent exactly the same state, so the phase $k$ is completely meaningless. 1s orbitals are states as well, so it is meaningless to talk about the phase of a 1s orbital.
However, if you have a state that is made up of two different states (like a molecular orbital), then the relative phases of the two constituent orbitals matter. ​That's why the bonding and antibonding MOs are different: it's because the two constituent parts don't have the same relative phase in the two MOs (in the bonding it's the same, in the antibonding they're opposite).

There are very few, and possibly no, quick and dirty ways to explain MO theory, in my opinion. However, if you have studied linear algebra before,* then the transition from AOs and MOs can be viewed as a change of basis. A basis refers to a minimal set of vectors, from which any arbitrary vector may be constructed through linear combination. If you have a vector $(a, b)$, you can express this as: $$\begin{pmatrix}a\\b\end{pmatrix} = a\begin{pmatrix}1\\0\end{pmatrix} + b\begin{pmatrix}0\\1\end{pmatrix}.$$
It doesn't matter what $a$ and $b$ are, you can always express $(a, b)$ as a linear combination of the two basis vectors. You can think of the basis vectors $(1, 0)$ and $(0,1)$ as being the 1s orbitals on both individual atoms.
The MOs correspond to using the basis $(0.5, 0.5)$ and $(0.5, -0.5)$.† Notice that this is still a valid basis, because we can still express any arbitrary vector $(a, b)$ as a linear combination of these:
$$\begin{pmatrix}a\\b\end{pmatrix} = (a+b)\begin{pmatrix}0.5\\0.5\end{pmatrix} + (a-b)\begin{pmatrix}0.5\\-0.5\end{pmatrix}.$$
If we had rejected the &quot;antibonding&quot; combination $(0.5, -0.5)$, and only taken the &quot;bonding&quot; combination $(0.5, 0.5)$, then we would not be able to form a valid basis. For example, we could not express the vector $(1, 2)$ as just a linear combination of $(0.5, 0.5)$.
From this point of view, AOs and MOs are simply different bases.‡ The existence of both the bonding and antibonding MO is not only logical, but also mandatory in order to form a complete basis.
The familiar AOs that you have studied (1s, 2s, 2p, ...) are simply a series of basis states for the hydrogen atom. In the $\ce{H2}$ molecule, there are two sets of basis states, one for each atom. The transition from &quot;2 sets of AOs&quot; to &quot;1 set of MOs&quot; is just a change of basis: we go from two sets of atomic bases to one single molecular basis.

The proper answer is, of course, to pick up a QM textbook and study it. By QM I do indeed mean quantum mechanics, not just MO theory. It is a long journey, and you may not necessarily be fully prepared for it right now, but it will reward you with a better understanding of how this works. MO theory fundamentally relies on quantum mechanics, and to learn MO theory without understanding quantum mechanics is what leads to confusion and questions like yours.

* The analogy I use above is slightly simplified, and might look slightly contrived, but the link between QM and linear algebra is actually very deep. QM is somewhat analogous to linear algebra but on an infinite-dimensional vector space (technically, a Hilbert space).
† Technically this should be $1/\sqrt{2}$, not $0.5$. In QM we are nearly always concerned with orthonormal bases, i.e. ones where each basis vector has a length of $1$ and forms a scalar product of $0$ with every other basis vector. (Mathematically, we have a basis $\{|i\rangle\}$ for which $\langle i | j \rangle = \delta_{ij}$.)
‡ &quot;Bases&quot; being the plural of &quot;basis&quot;.