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When discussing the Dirac notation of an operator, for example, let's just say we have the bit flip gate $X$ if we want to write this in the Dirac notation does that just mean writing it as follows?

$$X|psirangle=X(c_0|0rangle+c_1|1rangle)=c_0|1rangle+c_1|0rangle$$

The Dirac notation for the Pauli-$X$ gate is:

$$|1rangle langle0| + |0rangle langle1|.$$

Now you might be wondering where this comes from. The term you're looking for is outer product representation of the $X$ gate. It follows from the spectral decomposition theorem (check Nielsen & Chuang 10th edition, p. 72) which holds for all normal operators. The key point:

In terms of the outer product representation, this means that $M$ can be written as $M=sum_ilambda_i|iranglelangle i|$,where $lambda_i$ are the eigenvalues of $M$,$|irangle$ is an orthonormal basis for $V$, and each $|irangle$ an eigenvector of $M$ with eigenvalue $lambda_i$.

The eigenvectors of the Pauli-$X$ gate are $-|0rangle+|1rangle$ and $|0rangle+|1rangle$, and the corresponding eigenvalues are $-1$ and $+1$ cf. Wolfram Alpha. Normalize the eigenvectors to get an orthonormal basis for $X$ i.e. $frac0rangle+sqrt2,frac0rangle+sqrt2$. According the spectral decomposition theorem you can represent the $X$ gate as:

$$-1(frac0rangle+sqrt2)(frac-langle 0sqrt2) + 1(frac0rangle+sqrt2)(fracsqrt2)$$
$$=-frac12(|0ranglelangle0|-|0ranglelangle1|-|1ranglelangle0|+|1ranglelangle1|)+frac12(|0ranglelangle0|+|0ranglelangle1|+|1ranglelangle0|+|1ranglelangle1|)$$
$$=|1rangle langle0| + |0rangle langle1|$$

To convince you that this result is correct let's apply it on an arbitrary qubit state $c_0|0rangle+c_1 |1rangle$:

$$(|1rangle langle0| + |0rangle langle1|)(c_0|0rangle+c_1|1rangle)$$
$$=c_0|1ranglelangle0|0rangle+c_1|0ranglelangle 1|1rangle$$
$$=c_0 |1rangle + c_1 |0rangle$$

So yes, our result is correct and the bits were indeed flipped upon application of $X=|1rangle langle0| + |0rangle langle1|$ to $c_0|0rangle + c_1|1rangle$.The last step followed from the fact that $langle 0|0rangle$ and $langle 1|1rangle$ are both equal to $1$, as $|0rangle$ and $|1rangle$ are orthonormal vectors i.e. their inner product $langle psi|psirangle=1$ by definition.

We're done. As an exercise, find the outer product representation of the Pauli-$Z$ gate by yourself. And definitely, do go through the proof of the spectral theorem in Nielsen and Chung if time permits!

This might mean using the ketbra notation:

$$X = |1rangle langle0| + |0rangle langle1|$$

This notation describes the effect the operator has on the basis vectors: in this case $X$ converts $|0rangle$ into $|1rangle$ and vice versa.

A couple of other examples:

$$Z = |0rangle langle0| - |1rangle langle1|$$

$$operatornameCNOT = |0 ranglelangle0| otimes I + |1 ranglelangle 1| otimes X = |00rangle langle00| + |01rangle langle01| + |11rangle langle10| + |10rangle langle11|$$