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 induced 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, in this case, 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.

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.