The answer is no. To help keep our notation straight, assume that there is a finite group $H$ and normal subgroups $G_1$ and $G_2$ of $H$ such that $G_1 \cong G_2$ and such that we have short exact sequences $$1 \longrightarrow G_1 \longrightarrow H \stackrel{f}{\longrightarrow} A_4 \longrightarrow 1$$ and $$1 \longrightarrow G_2 \longrightarrow H \stackrel{g}{\longrightarrow} D_6 \longrightarrow 1.$$ Choose this $H$ such that its cardinality is as small as possible.

Define $$\overline{G}_1 = G_1 / G_1 \cap G_2 \quad \text{and} \quad \overline{G}_2 = G_2 / G_1 \cap G_2.$$ The homomorphism $f$ induces an isomorphism between $\overline{G}_2$ and a nontrivial normal subgroup of $A_4$, and the homomorphism $g$ induces an isomorphism between $\overline{G}_1$ and a nontrivial normal subgroup of $D_6$. What is more, since $G_1$ is isomorphic to $G_2$ the groups $\overline{G}_1$ and $\overline{G}_2$ have the same cardinality. Examining the nontrivial normal subgroups of $A_4$ and $D_6$, we see that the only possibility is that these subgroups are actually the entire groups, i.e. that $$\overline{G}_1 \cong D_6 \quad \text{and} \quad \overline{G}_2 \cong A_4.$$ This implies that we have short exact sequences $$1 \longrightarrow G_1 \cap G_2 \longrightarrow G_1 \longrightarrow D_6 \longrightarrow 1$$ and $$1 \longrightarrow G_1 \cap G_2 \longrightarrow G_2 \longrightarrow A_4 \longrightarrow 1.$$ We conclude that $G = G_1 = G_2$ is a group that of cardinality strictly smaller than $H$ with fitting into the desired exact sequences, contradicting the minimality of the cardinality of $H$.

(the original post only contained a bunch of observations about the problem, but in the comments Ian Agol pointed out that one could use them as above to give a negative answer)

The answer is no. To help keep our notation straight, assume that there is a finite group $H$ and normal subgroups $G_1$ and $G_2$ of $H$ such that $G_1 \cong G_2$ and such that we have short exact sequences $$1 \longrightarrow G_1 \longrightarrow H \stackrel{f}{\longrightarrow} A_4 \longrightarrow 1$$ and $$1 \longrightarrow G_2 \longrightarrow H \stackrel{g}{\longrightarrow} D_6 \longrightarrow 1.$$ Choose this $H$ such that its cardinality is as small as possible.

Define $$\overline{G}_1 = G_1 / G_1 \cap G_2 \quad \text{and} \quad \overline{G}_2 = G_2 / G_1 \cap G_2.$$ The homomorphism $f$ induces an isomorphism between $\overline{G}_2$ and a nontrivial normal subgroup of $A_4$, and the homomorphism $g$ induces an isomorphism between $\overline{G}_1$ and a nontrivial normal subgroup of $D_6$. What is more, since $G_1$ is isomorphic to $G_2$ the groups $\overline{G}_1$ and $\overline{G}_2$ have the same cardinality. Examining the nontrivial normal subgroups of $A_4$ and $D_6$, we see that the only possibility is that these subgroups are actually the entire groups, i.e. that $$\overline{G}_1 \cong D_6 \quad \text{and} \quad \overline{G}_2 \cong A_4.$$ This implies that we have short exact sequences $$1 \longrightarrow G_1 \cap G_2 \longrightarrow G_1 \longrightarrow D_6 \longrightarrow 1$$ and $$1 \longrightarrow G_1 \cap G_2 \longrightarrow G_2 \longrightarrow A_4 \longrightarrow 1.$$ We conclude that $G = G_1 = G_2$ is a group that of cardinality strictly smaller than $H$ that fits into the desired exact sequences, contradicting the minimality of the cardinality of $H$.

(the original post only contained a bunch of observations about the problem, but in the comments Ian Agol pointed out that one could use them as above to give a negative answer)

While I do not know the answer, I have some remarks that are too long for the comments section. To help keep our notation straight, assume that we have a group $H$ and normal subgroups $G_1$ and $G_2$ of $H$ such that $G_1 \cong G_2$ and such that we have short exact sequences $$1 \longrightarrow G_1 \longrightarrow H \stackrel{f}{\longrightarrow} A_4 \longrightarrow 1$$ and $$1 \longrightarrow G_2 \longrightarrow H \stackrel{g}{\longrightarrow} D_6 \longrightarrow 1.$$ Define $$\overline{H} = H / G_1 \cap G_2 \quad \text{and} \quad \overline{G}_1 = G_1 / G_1 \cap G_2 \quad \text{and} \quad \overline{G}_2 = G_2 / G_1 \cap G_2.$$

The homomorphism $f$ induces an isomorphism between $\overline{G}_2$ and a nontrivial normal subgroup of $A_4$, and the homomorphism $g$ induces an isomorphism between $\overline{G}_1$ and a nontrivial normal subgroup of $D_6$. What is more, since $G_1$ is isomorphic to $G_2$ the groups $\overline{G}_1$ and $\overline{G}_2$ have the same cardinality. Examining the nontrivial normal subgroups of $A_4$ and $D_6$, we see that the only possibility is that these subgroups are actually the entire groups, i.e. that $$\overline{G}_1 \cong D_6 \quad \text{and} \quad \overline{G}_2 \cong A_4.$$ We also see that $f(G_2) = A_4$ and $g(G_1) = D_6$. This implies that $\overline{H}$ is generated by $\overline{G}_1$ and $\overline{G}_2$. Since $\overline{G}_1$ and $\overline{G}_2$ are normal subgroups of $\overline{H}$ and $\overline{G}_1 \cap \overline{G}_2 = 1$, they have to commute with each other (this is a standard trick, see here). We conclude that $$\overline{H} = \overline{G}_1 \times \overline{G}_2 \cong A_4 \times D_6.$$ Unwinding this, we see that $$H / G_1 \cap G_2 \cong A_4 \times D_6.$$ I don't know what to do with this, but it seems like it should be key to either constructing an example or showing that one does not exist.

The answer is no. To help keep our notation straight, assume that there is a finite group $H$ and normal subgroups $G_1$ and $G_2$ of $H$ such that $G_1 \cong G_2$ and such that we have short exact sequences $$1 \longrightarrow G_1 \longrightarrow H \stackrel{f}{\longrightarrow} A_4 \longrightarrow 1$$ and $$1 \longrightarrow G_2 \longrightarrow H \stackrel{g}{\longrightarrow} D_6 \longrightarrow 1.$$ Choose this $H$ such that its cardinality is as small as possible.

Define $$\overline{G}_1 = G_1 / G_1 \cap G_2 \quad \text{and} \quad \overline{G}_2 = G_2 / G_1 \cap G_2.$$ The homomorphism $f$ induces an isomorphism between $\overline{G}_2$ and a nontrivial normal subgroup of $A_4$, and the homomorphism $g$ induces an isomorphism between $\overline{G}_1$ and a nontrivial normal subgroup of $D_6$. What is more, since $G_1$ is isomorphic to $G_2$ the groups $\overline{G}_1$ and $\overline{G}_2$ have the same cardinality. Examining the nontrivial normal subgroups of $A_4$ and $D_6$, we see that the only possibility is that these subgroups are actually the entire groups, i.e. that $$\overline{G}_1 \cong D_6 \quad \text{and} \quad \overline{G}_2 \cong A_4.$$ This implies that we have short exact sequences $$1 \longrightarrow G_1 \cap G_2 \longrightarrow G_1 \longrightarrow D_6 \longrightarrow 1$$ and $$1 \longrightarrow G_1 \cap G_2 \longrightarrow G_2 \longrightarrow A_4 \longrightarrow 1.$$ We conclude that $G = G_1 = G_2$ is a group that of cardinality strictly smaller than $H$ with fitting into the desired exact sequences, contradicting the minimality of the cardinality of $H$.

(the original post only contained a bunch of observations about the problem, but in the comments Ian Agol pointed out that one could use them as above to give a negative answer)