One of the most natural ways that these central extensions arise is in Clifford theory of group representations, a well-studied area. I limit my discussion to working with representations over algebraically closed fields $k$.

If we have an irreducible $kG$ module $V$ for a finite group $G$, and $G$ has a normal subgroup $N,$ there are two basic Clifford reductions. If ${\rm Res}^{G}_{N}(V)$ is not homogeneous ( ie not a sum of isomorphic irreducible $kN$-modules), then $V$ is induced from a proper subgroup $H$ of $G$ which contains $N$ (more can be said, but that is not so relevant here).

On the other hand, if ${\rm Res}^{G}_{N}(V) \cong eU$ for some irreducible $kN$-module $U$ and positive integer $e$, then while $U$ may not extend to an irreducible $kG$-module, it does extend to an irreducible $k{\hat G}$-module for some central extension ${\hat G}$ of $G$. The standard way of doing this is as follows: let $\sigma$ be the irreducible matrix representation of $N$ afforded by $U.$ Then in the case under consideration, for each $g \in G,$ the representation of $N$ given by $n \to \sigma(gng^{-1})$ is equivalent to the representation $\sigma,$ so there is an invertible matrix $T_{g}$ such that $\sigma(gng^{-1}) = T_{g}\sigma(n)T_{g}^{-1}$ for all $n \in N.$ However, the matrix $T_{g}$ is not quite unique : by Schur's Lemma, using the algebraic closure of $k,$ the matrix $T_{g}$ is unique up to a scalar multiple. Note then that for each $g,h \in G,$ we have $T_{gh} = \alpha(g,h)T_{g}T_{h}$ for some scalar $\alpha(g,h) \in k.$

Clifford's second reduction gives a factorization of $V$ as $U \otimes W$ as $k{\hat G}$-module, where $W$ is an $e$-dimensional irreducible $k\tilde{G/N}$-module, where $\tilde {G/N}$ is a finite cyclic central extension of $G/N.$ There are various normalizations required to enable us to deal with finite central extensions, and to deal with a $2$-cocycle of $G/N.$ Also, this allows us to choose each $\alpha(g,h)$ to a root of unity.

So one answer to this question is that the field necessary to obtain a non-trivial irreducible representation of a central extension of $G/N$ which may be of smaller dimension than any possessed by $G/N$ is explained by a $2$-cocycle of $G/N$ arising in Clifford's second reduction.

Later edit: Another way to look at this is that (in the second Clifford reduction case), the fact that $U$ is $G$-stable gives us an action of $G$ on ${\rm End}_{k}(U) \cong U \otimes U^{*}$, and the fact that all automorphisms of that matrix algebra are inner allows us to define an action of $G$ on $U$ "up to scalars".

In a slightly different direction, a somewhat related phenomenon is that ( for $p$ odd), the group ${\rm Sp}(2n,p)$ has "unreasonably small" irreducible characters of degree $\frac{p^{n} \pm 1}{2}$, known as the Weil characters (for reasons clear to specialists).
These can be obtained as follows: An extra-special group $E$ of order $p^{2n+1}$ and exponent $p$ has a faithful irreducible character $\theta$ of degree $p^{n}$ which is unique up to Galois conjugation, and is ${\rm Sp}(2n,p)$-stable. It turns out that $\theta$ does extend to the semi-direct product $E{\rm Sp}(2n,p)$. But on restricton back to ${\rm Sp}(2n,p)$, the character decomposes into two pieces, one for each eigenspace of the central involution of ${\rm Sp}(2n,p).$ These eigenspaces have dimensions $\frac{p^{n} \pm 1}{2}.$ Notice that when $p = 5$ and $n = 1,$ we obtain $2$ and $3$-dimensional representations of ${\rm Sp}(2,5) \cong {\rm SL}(2,5)$- the $2$-dimensional representation is faithful, and the $3$-dimensional representation has central kernel of order $2$. In general, the faithful Weil character of ${\rm Sp}(2n,p)$ has degree $\frac{p^{n}+ \varepsilon}{2}$ where the sign $\varepsilon$ is chosen to make this degree even.

Even later edit: Another interesting case occurs with $A_{7}.$ We have an isomorphism $A_{8} \cong {\rm GL}(4,2).$ The $4$-dimensional representation in characteristic $2$ restricts irreducibly to $A_{7}.$ This representation of $A_{7}$ does not lift to a characteristic $0$-representation of $A_{7}$ since $A_{7}$ has no non-trivial complex irreducible character of degree less than $6$. However the representation does lift to a complex representation of the double cover ${\hat A_{7}}$.

permutationrepresentation larger than the original. $\endgroup$ – Nick Gill Feb 26 '15 at 16:59