Given a single-qubit state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, is it fine to calculate something like:
$$\langle\Phi^+|\psi\rangle = \frac{1}{\sqrt{2}}(\alpha\langle0| + \beta\langle1|),$$
where $\langle\Phi^+| = \frac{1}{\sqrt{2}}(\langle00| + \langle11|)$?
I feel that it's okay, since if we have $\langle00|0\rangle=\langle0|\otimes\langle0|\otimes|0\rangle$, and by orthonormality, we can simplify to $\langle0|$.
Your two-qubit state is defined on (funnily enough) two qubits, let's call them $A$ and $B$. It is now important to know whether $|\psi\rangle$ is a state/projection/measurement of qubit $A$ or qubit $B$. Without at least that specification, the operation is highly ambiguous (although the two answers happen to be the same in the example you give).
For greater clarity (and mathematical correctness), you need to include identity, $I$, on the qubit that you're not projecting on. For example, either $$ \langle\Phi^+|(|\psi\rangle\otimes I)\qquad\text{or}\qquad \langle\Phi^+|(I\otimes|\psi\rangle). $$
