I think this may be easier to understand if you change the notation a bit. Instead of grouping the direct summands by their isomorphism type, just list them all without grouping. So we have two decompositions $V=\bigoplus S_m$ and $W=\bigoplus T_n$, where each $S_m$ and each $T_n$ is irreducible. Given an isomorphism $\varphi:V\to W$, let $\varphi_{mn}:S_m\to T_n$ be the composition of $\varphi$ with the inclusion $S_m\to V$ and the projection $W\to T_n$. By Schur's lemma, each $\varphi_{mn}$ is either an isomorphism or $0$.
Now since $\varphi$ is injective, for each $m$ there must exist some $n$ such that $\varphi_{mn}\neq 0$. Thus for each $m$, there exists some $n$ such that $\varphi_{mn}$ is an isomorphism, and hence $T_n\cong S_m$. Moreover, $\varphi_{mn}=0$ for all $n$ such that $T_n\not\cong S_m$. This means that image of the restriction of $\varphi$ to $S_m$ is contained in the direct sum of all the $T_n$'s which are isomorphic to $S_m$.
Now fix an irreducible representation $R$ and let $A\subseteq V$ be the direct sum of all the $S_m$'s that are isomorphic to $R$, and let $B$ be the direct sum of all the other $S_m$'s, so $V=A\oplus B$. Similarly, let $C\subseteq W$ be the direct sum of all the $T_n$'s that are isomorphic to $R$, and $D$ be the direct sum of all the other $T_n$'s, so $W=C\oplus D$. The discussion above shows that $\varphi(A)\subseteq C$ and $\varphi(B)\subseteq D$. Since $\varphi$ is surjective, we must have $\varphi(A)=C$ and $\varphi(B)=D$. Thus $\varphi$ gives an isomorphism from $A$ to $C$. It follows that the number of $S_m$'s which are isomorphic to $R$ is equal to the number of $T_n$'s which are isomorphic to $R$, which is exactly what we wanted to prove.
Note that you're right that, for instance, $\varphi(S_m)$ might not actually be equal to any of the $T_n$. For instance, if $G$ is trivial, this is just saying that if you have two bases for the vector space, you can have a vector in one basis that is not a scalar multiple of any single vector in the other basis. But $\varphi(S_m)$ is still isomorphic to one of the $T_n$. Moreover, $\varphi(A)$ is actually equal to $C$, or in the language of the question, $\varphi(V_i^{\oplus a_i})=W_j^{\oplus b_j}$ for some $j$. So while the individual irreducible summands might not map to individual irreducible summands, when you group together all the irreducible summands of a given isomorphism type, they map to the sum of all the irreducible summands of the same isomorphism type.
By definition, a simple module is nonzero. Hence its dimension, which equals its multiplicity in the decomposition, is positive. This guarantees that any irreducible representation does appear in it.
Best Answer
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 $1$.