This property of the tensor product has really nothing to do with its interpretation in quantum mechanics or your specific example. Given two vector spaces $U,V$ the tensorproduct $\phi: U \times V \to U \otimes V$ is characterized by the property (up to isomorphism), that for any bilinear map $b \colon U \times V \to W$ into some third vectorspace $W$, there exists a unique linear map $l \colon U \otimes V \to W$, such that $b = l\circ\phi$. Because the two projections from the direct product $U\times V \to U$ and $U \times V \to V$ are not bilinear, they do not yield a map $U\otimes V \to U$ or $U \otimes V \to V$.
In quantum mechanical parlance this is called entanglement. As you probably know in this case $U$ and $V$ are spaces of state and $U \otimes V$ is the space of state for the combined system. Notice that just because the projections do not yield maps $U \otimes V \to V$ this does not mean there are no such maps. If for instance $U = H_e$ is the state space of a relativistic electron and $V = H_\gamma$ the state space of a photon, there is a map $H_e \otimes H_\gamma \to H_e$ that describes the absorption of a photon by an electron. In this case one usually draws a feynman diagram with a wiggly line for the photon meeting the electron line.