Every character is the trace of a (finite-dimensional) representation; by Maschke's theorem, every such representation is a direct sum of irreducible ones, and therefore the original character is equal to the sum of the traces of these irreducible representations, possibly with multiplicity. The multiplicities are all positive integers, because the answer "how many times does a copy of the irrep $V$ appear inside the representation $W$?" is always a non-negative integer.
Conversely, if a class function $\psi$ is equal to $\sum_{i=1}^m c_i \chi_i$ where each $c_i$ is a positive integer, then $\psi$ is the trace of the representation $\bigoplus_{i=1}^k c_i \pi_i$, where we choose each $\pi_i$ to be an irreducible representation whose trace is $\chi_i$.
That was my attempt to answer the second question. To answer the first question: it is a character when it is a character.
Less facetiously, take the inner product of your class function with each of the irreducible characters, using the normalization if the inner product which makes the irreducible characters an orthonormal basis for $\ell^2(G)$. If each of the resulting numbers is a non-negative integer, this expresses your class function as a non-negative integral combination of irreducible characters.
The idea here is that an arbitrary algebraically closed field of characteristic $0$ can be endowed with a structure that emulates the behaviour of $\mathbb{C}$ together with the complex conjugation automorphism and the absolute value. By this I specifically mean the following
Theorem. Let $K$ be algebraically closed of characteristic $0$. Then there exists an involutive field automorphism $\iota \in \mathrm{Aut}_{\mathbb{Q}}K$ such that the fixed subfield $E:={}^{\iota}K$ is orderable by a certain total order $R$ and such that the structure $(E, +, \cdot, R)$ be a formally real field.
Sketch of proof: since it is of characteristic $0$, $K$ has a natural $\mathbb{Q}$-algebra structure; considering the extension $K/\mathbb{Q}$, let us fix a certain transcendence basis $B \subseteq K$.
It is known that free commutative monoids over any set (of arbitrary cardinality) are totally orderable (in the sense of admitting total orders compatible with the monoid structure) and that consequently polynomial rings (in arbitrarily many indeterminates) over (totally) ordered rings are (totally) orderable.
Hence in particular the integral domain $\mathbb{Q}[X_t]_{t \in B}$ is totally orderable and thus this order structure can be extended to the rational fraction field $\mathbb{Q}(X_t)_{t \in B}$, rendering it into a totally ordered field. Consider now a real closure $E$ of $\mathbb{Q}(X_t)_{t \in B}$ and furthermore an algebraic closure $F$ of $E$, which by the famous Euler-Lagrange theorem is actually a quaint quadratic extension of $E$ obtained by adjoining a square root of $-1_E$.
Clearly, $F$ and $K$ are by construction both algebraic closures of $\mathbb{Q}(X_t)_{t \in B}$ (to be specific with the details, $K$ is an algebraic closure of its subfield $\mathbb{Q}(B)$, the latter being canonically isomorphic to $\mathbb{Q}(X_t)_{t \in B}$), so they must be isomorphic $\mathbb{Q}(X_t)_{t \in B}$-algebras, via say an isomorphism $\varphi$.
By the general theory surrounding the Euler-Lagrange theorem, $F$ will naturally come equipped with a conjugation $\gamma$, which is none other than the automorphism fixing $E$ and taking $i$ to $-i$, where we have taken the liberty to denote by $i$ a certain fixed square root of $-1_E$. Transporting this entire structure via the isomorphism $\varphi$ introduced above is what establishes the existence of $\iota$ such that its fixed subfield be orderable with the structure of a really closed field. $\Box$
Having equipped $K$ with this structure, where we agree to denote the fixed subfield ${}^{\iota}K=E$, one can introduce the absolute value map on $K$ given by
$$| \bullet|: K \to K, \\ |z|=\sqrt{z \iota(z)}$$
since $z\iota(z)$, being fixed by $\iota$, will necessarily belong to $E$ and since really closed fields admit radicals of all orders (every positive element will have a unique positive root of order $n$ for any $n \in \mathbb{N}^*$).
The ''conjugation'' $\iota$ and the absolute value map $| \bullet |$ thus introduced exhibit exactly the same behaviour as the standard complex ones, and it is with this equipment that you can generalize results such as the one you discuss to the general setting of characteristic $0$ (with the proviso of algebraic closure, of course).
As a final remark, the ''conjugation'' introduced above is far from being a canonical object (somewhat unlike the case of complex conjugation, although ultimately one could argue that depending on the axiomatic system used to formalize mathematics not even the natural number $1$ is truly canonically fixed, in the sense of there being some arbitrariness behind its choice; but I digress with philosophical contemplations). However, the only important aspect is the existence of such structures, for they suffice to allow one to carry on the same kind of arguments and obtain the same type of inequalities/bounds as in the standard complex case.
Best Answer
I understand the 2D complex irrep best with quaternions.
Every quaternion looks like $a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$, in other words the formal sum of a scalar and a 3D vector. The product of any two of $\mathbf{i},\mathbf{j},\mathbf{k}$ in cyclic order is the third, and in reverse order is the opposite. If you check the multiplication table, then, you find if $\mathbf{u},\mathbf{v}$ are 3D vectors then the scalar and vector components (aka real and imaginary parts) of the quaternion product $\mathbf{uv}$ are $-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$ (since it is true for basis vectors). This implies the square roots of $-1$ are the unit vectors, and Euler's formula $\exp(\theta\mathbf{v})=\cos(\theta)+\sin(\theta)\mathbf{v}$ for unit vectors $\mathbf{v}$ follows. Furthermore, every quaternion has a polar form $r\exp(\theta\mathbf{v})$ with $r\ge0$ its magnitude, $\mathbf{v}$ a unit vectors, and $0\le\theta\le\pi$ convex (which is unique for nonreal quaternions, and $\theta=0$ or $\pi$ with $\mathbf{v}$ arbitrary for nonzero reals). If $\mathbf{u}$ is a unit vector, we can extend to an ordered orthonormal basis $\{1,\mathbf{u},\mathbf{v},\mathbf{w}\}$ of the quaternion algebra. If we write $L_q(x)=qx$ and $R_q(x)=xq$ with respect to this basis, where $q=\exp(\theta\mathbf{u})$, we see that the conjugation $C_q=L_q\circ R_q^{-1}$ given by $C_q(x)=qxq^{-1}$ restricts to a rotation of 3D vectors around the oriented axis by an angle of $2\theta$. In this way, $q\mapsto C_q$ is a double covering (i.e. $2$-to-$1$ group homomorphsim) $S^3\to\mathrm{SO}(3)$, where $S^3$ is the group of unit quaternions.
If $T\cong A_4$ is the tetrahedral group (of order $12$), its preimage under $S^3\to\mathrm{SO}(3)$ is the binary tetrahedral group $2T$ of order $24$. It forms the vertices of a $24$-cell, namely $\{\pm1,\pm\mathbf{i},\pm\mathbf{j},\pm\mathbf{k},\frac{1}{2}(\pm1\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k})\}$. The group is a semidirect product $2T=Q_8\rtimes C_3$ with $C_3$ generated by any $3$rd root of unity $\frac{1}{2}(-1\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k})$, with polar form $\omega=\exp(\frac{2\pi}{3}\frac{\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k}}{\sqrt{3}})$. Geometrically, we can place unit diameter $3$-spheres at all integer coordinates in 4D, which leaves just enough space to place unit diameter spheres centered at the hypercube centers to create an optimal sphere packing (the Hurwitz quaternions); the vertices of the spheres surrounding the central sphere forms the $24$-cell.
We can have $2T\curvearrowright\mathbb{H}$ act from the left and complex scalars from $\mathbb{C}=\{a+b\mathbf{i}\}$ act from the right (so it is a "right" vector space). Picking a basis $\{1,\mathbf{j}\}$ for $\mathbb{H}$ as a complex vector space allows us to write the right $\mathbb{C}$-linear transformations from $2T$ as $2\times2$ complex matrices. This is a 2D complex irrep of $2T$. We can check the character table to see tensoring this 2D rep with the nontrivial 1D reps gives the two other 2D irreps.
If we mod these $2\times2$ matrices by $3$ (the smallest modulus where $\pm1$ are distinct), we get $\mathrm{SU}_{1,1}(\mathbb{F}_{3^2})$, where $\mathbb{F}_{3^2}=\mathbb{F}_3(i)$, which is conjugate via the Cayley transform $[\begin{smallmatrix}1&-i\\1&i\end{smallmatrix}]$ to $\mathrm{SL}_2(\mathbb{F}_3)$. I elaborate a little more in my answer here (which shows the binary octahedral group $2O$ is a twisted version of $\mathrm{GL}_2\mathbb{F}_3$).