(I'll call the induced chart on the tangent bundle $(T \phi, TV)$ for quicker typing).
We have to prove that $T \phi \circ(fX) \circ \phi^{-1} : \phi[V] \subset \Bbb{R}^n \to T \phi[V] \subset \Bbb{R}^n \times \Bbb{R}^n$ is smooth, from the assumption that $X$ and $f$ are smooth, which means that $T \phi \circ X \circ \phi^{-1}$ and $f \circ \phi^{-1}$ are smooth maps (from the open subset $\phi[V] \subset \Bbb{R}^n$ into $\Bbb{R}^n \times \Bbb{R}^n$ and $\Bbb{R}$ respectively).
Well, now we just apply the definition of $T\phi$ and compute the compositions. Given any $\xi \in \phi[V]$, we have that
\begin{align}
(T \phi \circ (fX) \circ \phi^{-1})(\xi) &= T \phi \left( f(\phi^{-1}(\xi)) \cdot X(\phi^{-1}(\xi)) \right)
\end{align}
So, the vector $v$ we are considering is $f( \phi^{-1}(\xi)) \cdot X(\phi^{-1}(\xi)) \in T_{\phi^{-1}(\xi)}M$. If we apply $T\phi$ to this vector, the first $n$ entires will be the coordinates of the base point $\phi^{-1}(\xi)$, and the second $n$ entires will be the components of the vector (i.e apply $dx_i$ to this vector $v$). Hence,
\begin{align}
(T \phi \circ (fX) \circ \phi^{-1})(\xi) &= \left( \phi(\phi^{-1}(\xi)), \dots, dx_i[f( \phi^{-1}(\xi)) \cdot X(\phi^{-1}(\xi))] \dots \right) \in \Bbb{R}^n \times \underbrace{\Bbb{R}\times \dots \times \Bbb{R}}_{n\, \text{times}} \\
&= \left( \xi, \dots, (f \circ \phi^{-1})(\xi) \cdot dx_i (X \circ \phi^{-1})(\xi) \dots \right) \in \Bbb{R}^n \times \underbrace{\Bbb{R}\times \dots \times \Bbb{R}}_{n\, \text{times}}
\end{align}
(Here, I used the fact that $dx_i$ is linear in each tangent space, so that $dx_i(cv) = c \cdot dx_i(v)$)
Compare this with $(T\phi \circ X \circ \phi^{-1})(\xi) = \left( \xi, \dots, dx_i (X \circ \phi^{-1})(\xi) \dots \right) \in \Bbb{R}^n \times \underbrace{\Bbb{R}\times \dots \times \Bbb{R}}_{n\, \text{times}}$.
In other words, $(T \phi \circ (fX) \circ \phi^{-1})(\xi)$ is that $2n$-tuple whose first $n$ entries are simply $(\xi_1, \dots, \xi_n)$, and whose $(n+i)^{th}$ entry is obtained by multiplying the $(n+i)^{th}$ entry of $(T \phi \circ X \circ \phi^{-1})(\xi)$ and $(f \circ \phi^{-1})(\xi)$.
Stated as an equality of functions, If we write the coordinate functions as $(T \phi \circ (fX) \circ \phi^{-1})(\cdot) = (y_1(\cdot), \dots, y_{2n}(\cdot))$ then for $1 \leq i \leq n$, we have that
\begin{align}
y_i(\cdot) = (\text{id}_{\Bbb{R}^n})_i(\cdot) \quad \text{and} \quad y_{n+i}(\cdot) = (f \circ \phi^{-1})(\cdot) \cdot dx_i[X \circ \phi^{-1}(\cdot)]
\end{align}
Then, clearly each $y_i, y_{n+i}$ is a smooth function $\phi[V] \to \Bbb{R}$, hence the entire thing is smooth as a map from $\phi[V] \subset \Bbb{R}^n \to T\phi[V] \subset \Bbb{R}^{2n}$, which is what we needed to prove to show that $fX$ is a smooth vector field.
If you follow a similar reasoning for the sum $X+Y$ of vector fields, you'll find that
\begin{align}
y_i(\cdot) = (\text{id}_{\Bbb{R}^n})_i(\cdot) \quad \text{and} \quad y_{n+i}(\cdot) = dx_i[Y \circ \phi^{-1}(\cdot)] + dx_i[X \circ \phi^{-1}(\cdot)]
\end{align}
Hence, the sum of smooth vector fields is also smooth.
Best Answer
The composition $VW$ will be, in general, a non-linear differential operator, not a vector field.
For example, $\frac{\partial^2}{\partial x^2}$ is not a "derivation" in the sense that it doesn't verify the product rule of derivatives.
So, $VW$ is defined, of course, but it's not a vector field. This "product" is not an operation: it doesn't produce another vector field. It is a product in the algebra of differential operators, though.
On the other hand, the commutator of this product: $[V, W] = VW - WV$, called the Lie bracket of vector fields, is a linear differential operator: a vector field. This turns the module of vector fields into a type of noncommutative ring called Lie algebra. This is very important in the geometrical study of differential equations (the flow associated to $[V,W]$ is related with the flows of $V, W$ in an interesting way).