The reason behind the choice of $\lvert 0^{\otimes n}\rangle$ as reference state, found in many basic treatments of Grover's algorithm, is best understood by considering the technique that generalizes it: the so-called quantum amplitude amplification.
The goal of amplitude amplification is a very generic one: given some initial state
$\lvert\psi_{in}\rangle$, we want to transform it into a state that belongs to a given target subspace $\mathcal H_{target}$.
The initial state $\lvert\psi_{in}\rangle$ is assumed to be known, but $\mathcal H_{target}$ needs not be (and can indeed be seen as the goal of the algorithm).
This is consistent with what you have in the special case of Grover's algorithm with $\lvert\psi_{in}\rangle=\lvert+,\cdots,+\rangle=H^{\otimes n}\lvert 0,\cdots, 0\rangle$ and $\mathcal H_{target}=\mathbb C\lvert \psi_{target}\rangle$ one-dimensional and encoding the state that we are trying to find, and given only indirectly via oracular access to a function $f$ such that $f(x)=1$ iff $x=\lvert\psi_{target}\rangle$, and $f(x)=0$ otherwise.
In the general amplitude amplification scheme, the way we get from the initial to the target space is via repeated application of a pair of two reflections, $-S_t S_i$, where
$$S_i\equiv 2\lvert \psi_{in}\rangle\!\langle \psi_{in}\rvert-I, \\
S_t\equiv 2\lvert \psi_{t}\rangle\!\langle \psi_{t}\rvert-I,$$
and we used the notation $\lvert\psi_t\rangle\simeq\sum_{x\in\mathcal H_{target}}\lvert x\rangle$.
As it turns out, the product of two reflections amounts to a rotation in state space:
$$\newcommand{\ketbra}[1]{\lvert #1\rangle\!\langle #1\rvert}
R\equiv-S_t S_i=-4\ketbra{\psi_{in}}\psi_{t}\rangle\!\langle \psi_t\rvert
+2(\ketbra{\psi_{in}}+\ketbra{\psi_t})-I,$$
which brings the initial state $\lvert\psi_{in}\rangle$ closer to the target, as long as the initial overlap is not too big to begin with:
$$R\lvert\psi_{in}\rangle=(1-4\lvert\langle\psi_{in}\rvert\psi_{t}\rangle\rvert^2 )\lvert\psi_{in}\rangle+2\langle\psi_t\rvert\psi_{in}\rangle\lvert\psi_t\rangle,$$
$$\langle\psi_{in}\rvert R\lvert\psi_{in}\rangle=1-2\lvert\langle\psi_{in}\rvert\psi_{t}\rangle\rvert^2.$$
You can then verify how repeated applications of $R$ get you closer and closer to the target.
This shows how there is nothing special about the choice of $\lvert0\rangle$ often used: what is needed is a pair of reflections, one with respect to the initial state and the other with respect to the projector over the target state/space.
Why does the operator corresponding to the phase shift in Grover's algorithm correspond to $2\ketbra0-I$?
Let $\lvert\psi\rangle$ be any state, and define $S\equiv 2\ketbra\psi-I$. Then,
$$S\lvert\psi\rangle=2\lvert\psi\rangle-\lvert\psi\rangle=\lvert\psi\rangle,$$
while for any $\lvert\phi\rangle$ such that $\langle\phi\rvert\psi\rangle=0$,
$$S\lvert\phi\rangle=-\lvert\phi\rangle.$$
