Whenever you have a family over $B$, you get a morphism $B\rightarrow P$ such that the pullback of the universal family to $B$ recovers your family. Now consider any point $b\in B$. Then because your family is isotrivial, the composition $b\hookrightarrow B \rightarrow P$ must be mapped to the same point $p$, for every $b\in B$. This implies that the morphism $B\rightarrow P$ factors through a point $B\rightarrow p \rightarrow P$. But we know that the family over $B$ can be recovered by pulling back the universal family, so since the map factors through a point we see that the family you started with must be trivial. So if you have an isotrivial but non-trivial family, then that cannot be recovered by pulling back the universal family.
I think what you say about line bundles is correct. After all transition functions are all about $GL_n(\mathbb{C})$!
I'm not quite sure what you mean when you say that the data of a point in a stack is the single line bundle over a point along with the group of automorphisms $\mathbb{C}^*$. In any case, here is some intuition why considering a sheaf in groupoids (which gives stacks) rather than a sheaf in sets (which gives schemes/algebraic spaces) might help with dealing with isotrivial / locally trivial families. In the usual definition of a sheaf of sets $\mathcal{F}$ - say you have a covering $\{U_i\}$ of $X$ and sections $s_i \in \mathcal{F}(U_i)$, then asking for $s_i$ to glue together is to require $s_i|_{U_i\cap U_j} = s_j|_{U_i\cap U_j}$ for all $i,j$
When you consider a sheaf in groupoids, the latter condition is relaxed to asking for isomorphisms $s_i|_{U_i \cap U_j} \stackrel{\sim}{\rightarrow} s_j|_{U_i\cap U_j}$ (and some more conditions). This should remind you of how one glues together a non-trivial vector bundle!
This is really categorical nonsense to spell out.
$ \mathcal{M}_{g,1} $ is the category fibered in groupoids (cfig) over schemes over a field $ k $ whose objects consist of families $ \pi : X \rightarrow S $ where $ \pi $ is smooth proper with every geometric fiber a nonsingular genus $ g $ curve, along with a section $ s $ of $ \pi $. The morphisms are given by obvious commutative diagrams. $ \mathcal{M}_g $ is made up of the same data except the section. The morphism $ \mathcal{M}_{g,1} \rightarrow \mathcal{M}_g $ is also the obvious one, given by forgetting the section.
To see that this is the universal family, we need to show that every family in $ \mathcal{M}_g $ arises by pullback. So consider a family $ \pi : X \rightarrow S $ of curves as above (no section here). By the 2-Yoneda Lemma, there is a unique morphism $ h_{\pi} : \underline{S} \rightarrow \mathcal{M}_g $ inducing $ \pi $, where $ \underline{S} $ is the cfig associated to $ S $.
We claim that the fiber product $ \underline{S} \times_{\mathcal{M}_g} \mathcal{M}_{g,1} $ is $ \underline{X} $. By definition, an object of the fiber product is given by a compatible triple (over the same object $ T $ in the base category of schemes over $ k $), which means the following data:
(1) $ (T,g) $ an object of $ \underline{S} $, i.e. $ g : T \rightarrow S $ a morphism.
(2) $ \eta : Y \rightarrow T $ a family in $ \mathcal{M}_{g,1} (T) $ so that it comes with a section $ s $.
(3) $ \gamma $ an isomorphism given by $ \gamma : Y \rightarrow T \times_S X $ over $ T $.
The morphisms between objects in the fiber product are given by the obvious diagrams again.
Points (2) and (3) together imply that $ T \times_S X \rightarrow T $ has a section and giving a section of this map is equivalent to giving a map $ f : T \rightarrow X $ such that $ \pi \circ f = g $. All of this is clearly reversible: every map $ T \rightarrow X $ gives a compatible triple in the fiber product.
In summary, to specify the objects of the fiber product over $ T $ is equivalent to specifying morphisms $ T \rightarrow X $. So we proved that the fiber product is precisely $ \underline{X} $ and the morphism $ \underline{X} \rightarrow \underline{S} $ is $ \underline{\pi} $ by construction, showing that $ \mathcal{M}_{g,1} \rightarrow \mathcal{M}_g $ is indeed universal.
And yes, to answer one of your original questions, the moduli functor of smooth curves is not representable in the world of schemes, i.e. there does not exist a universal family (for the reasons of automorphisms as you allude to). The coarse moduli space just parametizes curves in the sense of closed points corresponding to curves. It doesn't come with a universal family.
Best Answer
Suppose $M$ represents your functor $\mathcal{M}_g$ and you have an isotrivial family $C\rightarrow B$ which is globally non-trivial. Then by definition, there exists a universal family $\mathcal{C}_g \rightarrow \mathcal{M}_g$ map $B\rightarrow M_g$ such that the pullback (fiber product) gives you $C\rightarrow B$.
Now if you take any $b\in B$, then under the composition $b\hookrightarrow B \rightarrow M_g$ corresponds to the family $C_b\rightarrow b$, where $C_b$ is the fiber of $C\rightarrow B$ over $b\in B$.
Since $C\rightarrow B$ is assumed to be isotrivial, $C_b \equiv C_{b'}$ for any $b,b' \in B$. Therefore $b,b'$ will map to the same point in $\mathcal{M}_g$. In particular we deduce that $B\rightarrow \mathcal{M}_g$ is the constant map, where every point maps to the point corresponding to the curve $C_b$.
However since $C\rightarrow B$ is the fiber product, it follows that $C\rightarrow B$ is the trivial family. This contradicts the assumption that $C\rightarrow B$ is non-trivial.