Zack, the coproduct in the category of associative algebras is quite different from the tensor product. For instance, the coproduct of two free associative algebras on one generator, $k\langle X\rangle$ and $k\langle Y\rangle$, is free on two generators, $k\langle X\rangle\amalg k\langle Y\rangle=k\langle X,Y\rangle$, while the tensor product is not $k\langle X\rangle\otimes k\langle Y\rangle=k[X,Y]$ since it is the commutative algebra of polynomials in two variables.

If you replace tensor product with *homotopy push-out* in your equation, and if all spaces are simply connected, then the result is true. In this case, the chain coalgebra $C_*(X)$ is the push-out $C_*(X_0)\oplus_{C_*(Y)}C_*(X_1)$ (this push-out is the same in the category of chain complexes and in the category of coalgebras). The cobar construction $\Omega$ is a left adjoint, so it preserves pushouts,
$\Omega C_*(X)=\Omega C_*(X_0)\amalg_{\Omega C_*(Y)}\Omega C_*(X_1)$. Now use everywhere the natural quasi-isomorphism $\Omega C_*(X)\simeq C_*(\Omega X)$ for simply connected spaces.

For non-simply connected spaces the result sketched in the previous paragraph is probably false. I think a counterexample can be found in the following way, although I haven't been brave enough to tackle the final step of the computation. Take $X_0=X_1=D^2$, $Y=X_0\cap X_1=S^1$ and $X=X_0\cup_YX_1=S^2$, in this case $C_*(\Omega S^2)\simeq k\langle x\rangle $ where $x$ has degree $1$ and trivial differential. The homotopy push-out $C_*(\Omega X_0)\amalg_{ C_*(\Omega Y)} C_*(\Omega X_1)$ is quasi-isomorphic to the suspension of $C_*(\Omega Y)$ in the category of augmented chain algebras, since $D^2$ is contractible. The augmented chain algebra $C_*(\Omega Y)$ is quasi-isomorphic to $k\langle y,z\rangle/(yz-1)$ concentrated in degree $0$. In order to compute its suspension, we have first to obtain a graded-free DG-resolution. This can be done by means of curved Koszul duality theory. I think that this algebraic suspension is going to be quite different to $k\langle x\rangle $.

In any case, in the non-simply connected case the natural replacement of your statement could maybe use the Baues-Tonks twisted cobar construction.