What is unique is the decomposition of a representation (over a field of characteristic not dividing $|G|$) into its isotypic components
$$V \cong \bigoplus_{\lambda} V_{\lambda}$$
where $\lambda$ denotes an irreducible representation and $V_{\lambda}$ denotes the sum of all subspaces of $V$ isomorphic to $\lambda$. For example, if $G = \mathbb{Z}_2$, then there are two isotypic components, the "even" one corresponding to the trivial representation and the "odd" one corresponding to the sign representation. More generally, if $G = \mathbb{Z}_n$ then there are $n$ isotypic components, one for each $n^{th}$ root of unity, which can be thought of as the eigenspaces of a generator of $G$ acting on $V$.
Each isotypic component $V_{\lambda}$ contains some number $k_{\lambda}$ of copies of the irreducible representation $\lambda$, which is uniquely determined as it must be $\frac{\dim V_{\lambda}}{\dim \lambda}$. The further decomposition of $V_{\lambda}$ into a direct sum of $k$ copies of $\lambda$, however is not unique; it is more or less equivalent to choosing a basis of the multiplicity space $\text{Hom}(\lambda, V)$. For example, if $V$ is trivial, decomposing $V$ as a direct sum of copies of the trivial representation is more or less equivalent to choosing a basis of $V$.
What Fulton and Harris mean by "unique" is the weaker statement that the isomorphism classes of irreducible representations $\lambda$ that occur are uniquely determined, as are their multiplicities $k_{\lambda}$. They do not mean that the subspaces in that decomposition are unique as subspaces, because they aren't in the presence of multiplicities higher than $1$.
There wasn't enough space for this as a comment so I've expanded into an answer. As I have said there is no formula for $\exp$ in the general situation. Indeed look at the formula for a linear Lie group:
$$ \exp(X) = I + X + \frac{1}{2}X^2 + \frac{1}{3!}X^3 + \cdots$$
What does $X^2$ mean? That is not something available in an abstract Lie algebra, so what's going on? The answer is that we have implicitly picked a representation. $X$ is really being used here to mean $\rho(X)$ and since $\rho(X)$ lives in $\mathfrak{gl}(V)$ which is a associative algebra as well as a Lie algebra $\rho(X)^2$ makes sense. Of course a representation is necessary to write something as matrices so this step is often overlooked when you only work with matrix Lie groups.
But a given Lie algebra does not just have one Lie group (indeed it doesn't even have to have just one matrix Lie group since we can pick different representations and get different groups that way). What it does have is a single simply connected Lie group. Such a group is often awkward to handle since, although it has all the same representations as the Lie algebra, they don't have to be faithful and only faithful representations can be visualised with matrices. As an example the Spin group has no faithful irreducible representations. So how can we find a way to represent elements?
The answer: don't bother. This is already hard enough for simple examples like the universal cover of $SL(2,\mathbb{R})$. However, we know how representations on this simply connected group work. You have written the formula down in your question: $\tilde{\rho}(\exp(X)) := \exp(\rho(X))$. Of course, not every element can be written as an exponential in general (unless your group is compact or nilpotent, for example) but it can be written as the product of exponentials. So you can piece together the whole action of the group that way.
So sacrificing our desire to have a "name" (i.e. a matrix) for each element. We can still understand how they interact, how they act on representations and how the exponential map behaves.
In short, $\gamma(t) := \exp(tX)$ defines a curve through $e$ called a 1-parameter subgroup whose derivative at $e$ is $X$. This curve is exactly the integral curve for the left-invariant (or right-invariant, both work) vector field generated by $X$. The ODE you referred to is simply conveying this idea. So instead of a formula we have a characterisation: it is the unique map that does this.
Best Answer
It should work for any "splitting field" (i.e. as long as the complex irreps are realizable via matrices over a smaller field). It is sufficent that your field $\mathbb{K}$ contains all $e$th roots of unity, where $e$ is the "exponent" of $G$, i.e. the LCM of orders of all $g\in G$. (Note $e$ may be smaller than $n=|G|$.) For smaller fields, as long as $\mathbb{K}[G]$ is still semisimple (i.e. as long as the characteristic of the field does not divide the order of $G$, so Maschke's theorem applies), the algebra $A=\mathbb{K}[G]$ may be written as a direct sum $\bigoplus n_iV_i$ of irreps $V_i$ with multiplicities $n_i$ (i.e. $nV:=\underbrace{V\oplus\cdots\oplus V}_n$ here).
Then the algebra $A$ according to the Artin-Wedderburn theorem is $A\cong \bigoplus \mathrm{End}(n_iV_i)$. Note it is relevant for this that $\hom(V_i,V_j)=0$ for distinct irreps ($i\ne j$). Then, in turn, $\mathrm{End}(nV)$ may be written as $n\times n$ matrices with entries that are themselves from the algebra $D=\mathrm{End}(V)$. When $\mathbb{K}$ is a splitting field, Schur's lemma says every element of $D$ is an isomorphism (of $V$) hence $D=\mathbb{K}$. However, otherwise, $D$ is simply a division algebra over $\mathbb{K}$ (which could be a field extension of $\mathbb{K}$, or a quaternion algebra, or whatever). Thus
$$ A\cong \bigoplus_i M_{n_i}(D_i) $$
where $D_i=\mathrm{End}(V_i)$. (I am sweeping some stuff about "opposite algebras" under the rug.)