Suppose we have two Riemannian manifolds $(B,g_B)$ and $(F,g_F)$ and consider their product $C:=B\times F$ endowed with the metric tensor $g_C:=-\pi_B^*(g_B)+\pi_F^*(g_F)$, where $\pi_B$ and $\pi_F$ denote the projections on $B$ and $F$ respectively and $^*$ is the pullback.
If I want to write the Ricci tensor of $C$, if $g$ was the usual product metric then that tensor would be just the sum of the two Ricci's on $B$ and $F$, but in this case the sign $-$ in front of $\pi_B^*(g_B)$ should change the sign of the Ricci of $B$: if $\xi:=x+v$, where $x\in T_bB$ and $v\in T_fF$ for $b\in B$ and $f\in F$, I would have$$\textrm{Ric}_{(b,f)}^C(\xi,\xi)=-\textrm{Ric}^B_b(x,x)+\textrm{Ric}^F_f(v,v).$$
What confuses me is the fact that if I change the sign in a metric of a Riemannian manifold the curvature tensors do not change (since the Riemann tensor is made of Christoffel symbols in coordinates which are products of the matrices of the metric) but here the Ricci does. Does this come from the fact that the metric contraction wrt the $g_C$?
If I have a smooth function $f:X\rightarrow\mathbb{R}$ than the Hessian of $f$ would be the same of $B$ but with the same sign?
P.s. I did everything not in coordinates, using, given a frame $\{E_i\}_i$ on $C$,$$\textrm{Ric}^C(X,Y)=\sum_ig_C(E_i,E_i)\,g_C(\textrm{R}^C_{XE_i}Y,E_i),$$following O'Neill computations in chapter $7$ of his book "Semi-Riemannian geometry".