How can the position representation make sense with compatibility of addition? (Dirac Notation) According to the definition of complex inner product is that: $$⟨\psi|\phi_{1} + \phi_{2}⟩ = \left<\psi|\phi_{1}\right> + \left< \psi| \phi_{2} \right>, \forall \psi, \phi.$$
This implies that with the position representation of the wavefunction, which states that $\left< r| \psi \right> = \psi(r)$. It leads to $ \psi(x+y) = \left< x+y | \psi \right> = \left< x | \psi \right> + \left< y | \psi \right> = \psi(x) + \psi(y)$. I think the [abuse of] notation is just making me confused about the ideas, but this doesn't seem right. With $x, y$ being continuous variables, what does this mean/represent? Is it valid to write? Is $\psi(x+y)$ even sensible?
 A: You're right that you're being tripped up by notation.  $x$ and $y$ are $\mathbb R$-valued labels for the (generalized) eigenvectors of the position operator, i.e. $|x\rangle$ is a vector such that $\hat X|x\rangle = x|x\rangle$.  It is certainly not the case that the vector labeled by $x+y$  is equal to the sum of the vectors whose labels are $x$ and $y$, respectively. In other words,
$$|x+y\rangle \neq |x\rangle + |y\rangle$$
On the other hand, if $|\psi_1\rangle$ and $|\psi_2\rangle$ are vectors, then we do have that
$$\bigg(\langle \psi_1| + \langle \psi_2| \bigg) |\phi\rangle = \langle \psi_1|\phi\rangle + \langle \psi_2 |\phi\rangle$$
As a general rule, it is not a good idea to write $|\psi+\phi\rangle = |\psi\rangle+|\phi\rangle$ when using Dirac notation, because it gives rise to precisely the question you raise.  If we avoid Dirac notation and identify $|x\rangle\equiv \phi_x$, then the issue becomes resolved quickly:
$$\psi(x+y) = \langle \phi_{x+y},\psi\rangle \color{red}{\overset{!}{\neq}} \langle \phi_x+\phi_y,\psi\rangle = \langle \phi_x,\psi\rangle + \langle \phi_y,\psi\rangle = \psi(x)+\psi(y)$$
