This is essentially a "down to earth" version of what David Ben-Zvi is saying. The object/example produced below essentially matches with what David is suggesting. I am just producing it "geometrically" to convince anyone interested that it is not a particularly strange beast and one doesn't need to know about dg-models or Koszul duality.

Let $\mathbb{G}_m$ act on itself by left multiplication. Let $a\colon \mathbb{G}_m\to pt$ be the canonical map. Consider the "silly" diagram (the functor $a_*$ is derived):

$\require{AMScd}$
\begin{CD}
D^b_{\mathbb{G}_m}(\mathbb{G}_m) @>{For}>> D^b(\mathbb{G}_m)\\
@VV{a_*}V @VV{a_*}V\\
D^b_{\mathbb{G}_m}(pt) @>{For}>> D^b(pt)
\end{CD}

This diagram commutes up to canonical isomorphism (essentially by construction/definition of the equivariant derived category). Write $\underline{k}$ for the (equivariant) constant sheaf on $\mathbb{G}_m$ (i.e., monoidal unit). Clearly,

$$ For(a_*\underline{k})= a_*For(\underline{k}) = H^*(\mathbb{G}_m)= k\oplus k[-1] $$

However, $a_*\underline{k}$ is indecomposable, because 

$$Hom(k, a_*\underline{k}) = H^*_{\mathbb{G}_m}(\mathbb{G}_m) = k$$

Here, the first "$k$" on the left hand side is the constant sheaf in $D^b_{\mathbb{G}_m}(pt)$ (this object generates the whole triangulated category, so it will detect decompositions), and as before the "$k$" on the far right hand side is just the field of coefficients.

The point, from the Koszul dual/algebraic perspective, is that any situation with torsion (i.e., where you can't obtain ordinary cohomology from equivariant cohomology) will lead to this. The characteristic 2 example I gave earlier is also "torsion phenomenon" but in a different sense.