Yes, it's correct, assuming to be precise that what you mean by $``R_1 + R_2"$ is
$$
R(X,Y)Z = (R_1(X_1, Y_1) Z_1, R_2(X_2, Y_2)Z_2).
$$
I imagine there's some elegant way to prove this, but it certainly works to just do a bunch of silly calculations; in particular, you can show that for product vector fields $(X_1, X_2)$ and $(Y_1, Y_2)$, the connection of $M$ is given by
$$
\nabla_{(X_1, X_2)} (Y_1, Y_2) = (\nabla^{M_1}_{X_1} Y_1, \nabla^{M_2}_{X_2} Y_2)
$$
using, for instance, the Koszul formula (I think you can also just check directly that the right side satisfies the properties of a Levi-Civita connection on $M$); then you can just use the standard defining formula for $R$
$$
R(X,Y) = \nabla_X\nabla_Y - \nabla_Y \nabla_X - \nabla_{[X,Y]}
$$
and it seems to just pop out.
According to the introduction paragraph of section 22.6. that you linked, they say that $R(X,Y)Z$ is the curvature of the given connection $\nabla$ which can by any connection.
Best Answer
By definition $$ R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z $$ which is a vector. Thus, $R$ takes in $3$ vectors and outputs a vector.