# Levi Civita connection on product of Riemannian Manifold

I'm trying to prove that for a product Riemannian manifold, say $$M=M_1\times M_2$$ with product metric one has $$\nabla_{X_{1}+X_{2}}^{M}\left(Y_{1}+Y_{2}\right)=\nabla_{X_{1}}^{M_{1}}\left(Y_{1}\right)+\nabla_{X_{2}}^{M_{2}}\left(Y_{2}\right)$$ for $$X_i,Y_i\in TM_i$$ via Koszul formula. So i choose $$Z=Z_1+Z_2$$ and i get $$$$2g(\nabla^M _{X_1+X_2} Y_1+Y_2,Z)=2g(\nabla ^{M_1}_{X_1}Y_1,Z_1)+2g(\nabla ^{M_2}_{X_2}Y_2,Z_2)+ X_2g(Y_1,Z_1)+X_1g(Y_2,Z_2)+Y_1g(X_2,Z_2)+Y_2g(X_1,Z_1)-Z_1g(X_2,Y_2)-Z_2g(X_1,Y_1)$$$$

I'd like that those therms cancels each out so the first two therms give the thesis, right? but i cant see how this happens

Take, for example, $$X_2 g(Y_1,Z_1)$$. $$\require{cancel}g(Y_1,Z_1)=g_1(Y_1,Z_1)+\cancel{g_2(Y_1,Z_1)}$$ is constant in the second coordinate $$p_2\in M_2$$, and $$X_2$$ is a derivation on $$M_2$$. So $$X_2 g(Y_1,Z_1)=0$$. Similarly the other terms.