I want to know if $$F_*(X\cdot Y)=F_*X\cdot F_*Y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ where $F:M\rightarrow N$ is smooth and $X,Y\in \mathcal{X}(M)$. The operation $\cdot$ (which I will omit from now on) is the composition of vector fields as derivations: $X\cdot Y(f):=X(Y(f))$, where $f\in C^\infty(M)$.
I tried the following:
Let $g\in C^\infty(N)$. Then:
$$ F_*(XY)(g) = XY(g\circ F)= X\Big( Y(g\circ F) \Big) $$
On the other hand
$$ (F_*X\ F_*Y )(g) = F_*X (F_*Y (g)) = F_*X\Big( Y(g\circ F) \Big) = X\Big( Y(g\circ F) \circ F \Big) $$
So it seems that the proposition is false after all. However, it seems that
$$ (X\ F_*Y )(g) = X (F_*Y (g)) = X\Big( Y(g\circ F) \Big) = F_*(XY)(g) $$ So that $$X\ F_*Y=F_*(XY)$$
I wonder if this is true.
Context:
I was hoping (1) to be true because I was originally trying to prove the following:
Let $F:G\rightarrow H$ be a Lie group homomorphism. Then, $F_*$ is a Lie algebra homomorphism.
So I want to show that $$[F_*X,F_*Y]=F_*[X,Y]$$ and this is a 3-step proof if $(1)$ holds.