Irreducible reps and characters of $G \rtimes A$

I might as well turn my comment into an answer. I will just write $GA$ for the semidirect product ( with the normal subgroup being $G$). Clifford's theorem outlines a procedure for computing the character table, but it is usually not straightforward in practice.

Firstly, we need to compute a set of orbit representatives for the complex irreducible characters of $G$ under the action of $A$.

Then, if the irreducible character $\chi$ of $G$ (one orbit representative) has stabilizer $A_{\chi}$, we need to find all irreducible characters of $GA_{\chi}$ whose restriction to $G$ is a multiple of $\chi$. In practice, this is usually the most difficult step, and often involves working with central extensions of $A_{\chi}$ and the determination of $2$-cocycles.

Once the irreducible characters of the required type for $GA_{\chi}$ have been determined, we induce them all from $GA_{\chi}$ to $GA$.

Doing this for each orbit gives all irreducible characters of $G$ exactly once.

When $A$ is cyclic, this procedure simplifies quite a bit, and even more so when $A$ has prime order.

For example, in the case you are looking at, of $G \langle t \rangle$ with $t$ of order $2$, it is necessary to determine the orbits of $t$ on the irreducible characters of $G = M_{12}$.

Once these are found, if the irreducible character $\chi$ of $M_{12}$ lies in an orbit of length $2$, then ${\rm Ind}_{G}^{G\langle t \rangle}(\chi)$ is an irreducible character of $G\langle t \rangle$ which vanishes identically outside $G$ and agrees with $\chi + \chi^{t}$ on $G$.

However, if an irreducible character $\chi$ lies in an orbit of length $1$, so is $\langle t \rangle$ invariant, there are exactly two irreducible extensions of $\chi$ to $G \langle t \rangle$, say ${\tilde \chi}$ and $\lambda {\tilde \chi}$, which differ by multiplication by the unique linear character $\lambda$ of $G\langle t \rangle$ with kernel $G$ (so that $\lambda(g) = -1$ for all $g \in G \backslash \langle t \rangle$).

There is work to do in determining the values of ${\tilde \chi}$ outside $G$.

(Of course, an alternative strategy in this case is to look up the character table in the Atlas!).

Later edit: I see that Derek Holt's comment, made while I was writing, has indicated the same procedures in the special case under consideration.

Here is a more conceptual approach to Clifford theory. Let me work with a slightly more general setup: namely, suppose we have a short exact sequence

$$1 \to N \to G \to H \to 1$$

of finite groups, which does not necessarily split. What does the representation theory of $G$ look like, in terms of $N$ and $H$? This question can be answered categorically as follows. The idea is to think of $G$ as the homotopy quotient $N//H$ of $N$ with respect to a (categorical) action of $H$. This gives an equivalence

$$\text{Rep}(G) \cong \text{Rep}(N)^H$$

where $(-)^H$ denotes taking homotopy fixed points. As a category,

$$\text{Rep}(N) \cong \prod_{\hat{N}} \text{Vect}$$

where $\hat{N}$ denotes the set of isomorphism classes of irreducible representations of $N$. The action of $H$ on this category breaks up based on the orbits of the action of $H$ on $\hat{N}$, so from now on we restrict our attention to a single orbit. If $\chi$ is an irrep in this orbit then we denote this orbit by $H\chi$.

Let $H_{\chi}$ denote the stabilizer of $H$ acting on $\chi \in \hat{N}$. Then we have an equivalence

$$\left( \prod_{H \chi} \text{Vect} \right)^H \cong \text{Vect}^{H_{\chi}}$$

where this copy of $\text{Vect}$ is spanned by $\chi$. So now it suffices to understand the homotopy fixed points of $H_{\chi}$ acting on a single copy of $\text{Vect}$.

Actions of $H_{\chi}$ on $\text{Vect}$ are classified by classes in $H^2(H_{\chi}, \mathbb{C}^{\times})$. Given a $2$-cocycle $c$ representing such a class, the category of homotopy fixed points is the category of projective representations of $H_{\chi}$ with $2$-cocycle $c$. I don't know if there's an easy recipe for computing these $2$-cocycles; Geoff Robinson also indicates that this is the hard step.

Anyway, once the irreducible projective representations of $H_{\chi}$ with $2$-cocycle $c$ have been determined, passing through the equivalences we've been using reproduces Geoff Robinson's answer: irreps of $G$ are classified by projective irreps of the $H_{\chi}$ with appropriate cocycles. From here the nicest thing that can happen is that

$$H^2(H_{\chi}, \mathbb{C}^{\times}) = 0$$

for all $\chi$, which in fact occurs in this example: $H_{\chi}$ is either trivial or cyclic of order $2$. So as Geoff Robinson and Derek Holt indicate, in this case it suffices to understand the orbits of the action of $\mathbb{Z}_2$ on the set of irreps of $M_{12}$.

For some blog references see here, here, here, here, and here, but the sequence isn't done yet. For references in the literature I think Ganter and Kapranov is relevant but I haven't looked at it in detail.