It is the second, $\psi(x) = \langle x|\psi\rangle$ which is correct. The first, if $x$ is the position operator, is just the position operator acting on the state $|\psi\rangle$. The abstract state $|\psi\rangle$ can be expanded in any basis, using a completion relation: $$|\psi\rangle = \underbrace{\sum_i |i\rangle \langle i|}_{1~=~identity}\psi\rangle$$ (where the sum can mean sum or integral, depending on this situation) using some complete basis ${|i\rangle}$. An example of this is the position basis, ${|x\rangle}$, from which we have $$|\psi\rangle = \int dx~ |x\rangle \langle x|\psi\rangle$$ This shows us that the wavefunction $\psi(x) = \langle x|\psi\rangle$, is the coefficient from the expansion of the state $|\psi\rangle$ in the position basis $|x\rangle$, at position $x$.
Edit to respond to a question asked by <PERSON> in response to <PERSON>'s answer: "But doesn't an operator acting on the state project the state onto the eigenstates of the operator and is that the same as representing the state in a new basis?"
As I have shown above, the "projection" you are talking about is given by applying the identity in terms of the position basis $\mathbf{1=\int dx~|x\rangle\langle x|}$.
Let's see what happens when we apply $\hat{x}$ instead: Just like state vectors can be expanded in a complete basis, so too can operators. 66
In general, if we take our complete basis ${|i\rangle}$ we can write the operator $\hat{A}$ in terms of matrix elements $$\hat{A} = \sum_{i,j} |i\rangle\langle i |\hat{A}|j\rangle\langle j |$$ where the matrix elements are $A_{ij} = \langle i |\hat{A}|j\rangle$. For the case of the position operator $\hat{x}$ and using the position basis ${|x\rangle}$ $$\hat{x} = \int dx~dy~ |x\rangle\langle x |\hat{x}|y\rangle\langle y |\ = \int dx~ x~|x\rangle\langle x |$$ that is, it is diagonal in the position basis (obviously). Using this we see $$\hat{x}|\psi\rangle = \int dx~ x~|x\rangle\langle x |\psi\rangle$$ So we see that application of $\hat{x}$ can be thought of as a kind of projection, weighted by $x$. So we see that it is not the same as representing the state in the $x$ basis, which is actually $$|\psi\rangle = \int dx~ |x\rangle \langle x|\psi\rangle$$ as above. 66