This question is geared towards the experts, so I will only briefly gloss the definitions. Everything I say is in the category of finite-dimensional smooth manifolds, and whenever I say "$\mathbb Z$-graded" I'm implicitly using the Koszul or "super" rule for signs. Recall that a vector bundle $A\to X$ determines a $\mathbb Z$-graded commutative algebra whose degree-$k$ part is $\Gamma(\wedge^k A^*)$. A Lie algebroid is a vector bundle $A\to X$ along with a degree-one square-zero derivation (Koszul rule!) on $\Gamma(\wedge^k A^*)$, i.e. it makes $\Gamma(\wedge^k A^*)$ into a complex and in fact a differential graded algebra. Lie algebroid cohomology is the homology of this complex (closed sections mod exact sections). (The "co", as far as I can tell, comes from the typical example: for any $X$, the tangent bundle ${\rm T}X \to X$ is a Lie algebroid via the de Rham $d$, and the homology of the complex $\Gamma(\wedge^k {\rm T}^* X)$ is de Rham cohomology of $X$. Or maybe the etymology is different. The functor from Lie algebroids to their cohomology is contravarian, so maybe that's it.) A morphism of Lie algebroids is a morphism of vector bundles that induces a morphism of the corresponding complexes.

A VBLA is a "vector bundle in the category of Lie algebroids". I.e. it is a pair of Lie algebroids $D \to B$ and $A\to X$, a morphism of algebroids $\{D\to B\} \to \{A\to X\}$, and maps $0: \{A\to X\} \to \{D\to B\}$ and $+ : \{D\to B\} \times_{\{A\to X\}} \{D\to B\} \to \{D\to B\}$ satisfying the conditions you would think of if you write down the words "vector bundle" in categorical language. (Equivalently, the induced maps on manifolds $B \to X$ and $D\to A$ are vector bundles, and there's some compatibility conditions.) For details, peruse the papers by Mehta and Garcia-Saz, or by Mackenzie.

In particular, a VBLA is a morphism of algebroids, and so induces a map on cohomology. My question is:

What "geometric" conditions on the VBLA are equivalent to the map on cohomology being an isomorphism? For example, does every VBLA induce an isomorphism on cohomology? If not, is there a way to check whether the cohomologies are isomorphic without computing the full cohomologies of both?

I'm realizing that I don't know enough examples (that I can actually compute) to gather up intuition. And the answer must be known by the experts (I'm guessing that the answer is known much more generally than just for VBLAs?). As with any mathoverflow question, I'm perfectly happy with an answer consisting entirely of a good reference.