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.
No, that the representations are isomorphic means they have the same character $\chi$ so it must be $F$-valued, there exists some representation over $F$ such that $tr \ \Pi = d \chi$ but sometimes we need a larger field to get $d=1$.
That's the difference between the extension $K$ (of $\Bbb{Q}$) generated by the character table and a splitting field for $G$ : a finite extension (of $K$) where all the irreducible representations can be realized.
The character table of the quaternions $Q_8$ is in $\Bbb{Q}$ but some of the representations need fields like $\Bbb{Q}(i)$ (or $\Bbb{Q}(\sqrt{-1-m^2})$ to be realized.
Best Answer
Sorry, my previous answer was wrong. I can't delete it, because it is an accepted answer, but I can edit it and completely change its contents!
I was confusing isomorphism classes of submodules with the submodules themselves.
According to Theorem 13.3, Chapter 5 of Huppert's German book "Endliche Gruppen I", in your situation. we have $W =\oplus_{i=1}^k U^{\sigma_i}$, where $U$ is an irreducible $LG$-submdule of $W$, and $\{\sigma_1,\ldots,\sigma_k\}$ is a (possibly proper) subset of ${\rm Gal}(L,F)$. (I haven't found a reference for this in English yet.)
I guess the answer to your question depends on what exactly you mean by "each $W_i$ is the direct sum of an entire ${\rm Gal}(L/K)$-orbit of irreps of $G$ over $L$".