I don't fully understand the hostility to thinking of motives in terms of compatible systems. It's true that a motive is not just a compatible system of realizations (or else we would call it a "compatible system" and not a motive) but it's also true that studying compatible systems has always been a good way to get intuition for many aspects of the theory of motives. Maybe a better thing to say is that a motive should be a *geometric recipe* which produces a realization in any existing or future cohomology theory.

You asked how to reconcile this perspective on 1-motives with the definition of 1-motives. I guess this means how to derive a compatible system of realizations from a map $X \to G$.

The realization of the 1-motive $X \to G$ in any cohomology theory should be the extension of $H^1(G)$ by $H^1(X^\vee)$. (More properly I guess this is the dual of the realization).

You ask if this is an extension of $H^1(A)$. The group $G$ is always an extension of an abelian variety $A$ by a torus $T$, and $H^1(G)$ is an extension of $H^1(T)$ by $H^1(A)$. So we see that we are potentially extending $H^1(A)$ in both directions.

Which extension? The idea is we should morally get the $H^1$ of the quotient space $G/X$. Since this space is not very well-behaved, we can simulate this, in an arbitrary cohomology theory, as follows:

We observe that any class in $H^1(G)$ (ideally, completely represented by a torsor) is multiplicative in the sense that its pullback to $G \times G$ under the multiplication map is equal to the sum of its pullbacks under the two projections. For torsors, we can fix an isomorphism of torsors realizing this.

An element of $H^1( X \to G)$ is a class in $H^1(G)$ together with a trivialization of its pullback to $H^1(X)$ which is multiplicative in the sense that the pullback of this trivialization to $X \times X$ along the multiplication is equal to the sum of its pullbacks under the two projections (using the above isomorphism to define this sum).

You can follow this recipe in any theory and get a concrete description of the extension.

Furthermore, in every reasonable cohomology theory you will get a clue to the idea that this construction produces *all* the extensions with "geometric origin". For example, in $\ell$-adic cohomology theories, you will see that extensions of $H^1(G)$ by $X^\vee$ correspond to a Selmer group, and extensions arising from this construction are the image of the map from the group of rational points to the Selmer group. So as long as the Tate-Shafarevich group is finite, there will not be room to fit any more extensions.