Let $E,F_1,F_2,G$ be Banach spaces and $U \subseteq E$ be an open subset. Let $f_j : U \longrightarrow F_j$ be $C^r$ maps and $B : F_1 \times F_2 \longrightarrow G$ be a bounded bilinear map. Then $g : x \mapsto B(f_1(x),f_2(x))$ is $C^r.$ Also for $k=1,2,\cdots,r$ we have $$D^kg(x) (h_1,h_2,\cdots,h_k) = \sum\limits_{j=0}^{k} \binom {k} {j} B\left (D^jf_1(x) (h_1,h_2,\cdots,h_j) ,D^{k-j}f_2(x) (h_{j+1},h_{j+2}, \cdots, h_k) \right ).$$
I have managed to prove that $g$ is $C^1$ and for any $h \in E$ we have $$Dg(x) (h) = B \left (f_1(x),Df_2(x)(h) \right ) + B \left (Df_1(x)(h),f_2(x) \right ).$$
Now how do I prove that $g$ is $C^k$ for $k \gt 1\ $? How do I prove first that $Dg \in C^1\ $? Here $Dg : U \longrightarrow \mathcal L(E,G).$ Can $Dg$ be written as a composition of Frechet differentiable functions? Can anybody please give me some suggestion? Then I will try to prove it for any $k \gt 1.$
Thanks for your time.
EDIT $:$ What I get is that $$D^2 g(x) (h_1,h_2) = B \left (f_1(x),D^2f_2 (x) (h_1,h_2) \right ) + B \left (Df_1(x)(h_1), Df_2 (x) (h_2) \right ) + B \left (Df_1(x) (h_2), Df_2 (x) (h_1) \right ) + B\left (D^2f_1(x) (h_1,h_2),f_2(x) \right).$$ But I don't know why it has been claimed in the proposition that $$B \left (Df_1(x)(h_1), Df_2 (x) (h_2) \right ) = B \left (Df_1(x) (h_2), Df_2 (x) (h_1) \right ).$$
I won't prove the $k^{th}$ product rule formula, because it is already tedious enough to do in one-dimension (though if you keep your cool, it is just a matter of induction and carefully playing around with some binomial coefficient identities... try to mimic the one-dimensional proof as much as possible).
As a general remark, whenever you want to prove certain maps are of class $C^k$, very useful are the canonical injections and projections $\iota_i:V_i\to V_1\times V_2$, and the canonical projections $\pi_i:V_1\times V_2\to V_i$ (for example $\iota_1(x):=(x,0)$ and $\pi_1(x,y):=x$). These are bounded linear maps hence $C^{\infty}$ (their second derivative vanishes).
For your particular question, there are a few ways of arguing things, depending on how much stuff you've already proven. For example, if you've already proven that the composition of $C^k$ functions is $C^k$, more precisely:
then this is quite simple. Here's how we apply in your specific case: we can write \begin{align} g= B\circ \left(\iota_1\circ f_1 + \iota_2\circ f_2\right) \end{align} Just to see that everything makes sense, observe that $f_1:U\to F_1$, $\iota_1:F_1\to F_1\times F_2$ so the composition is $\iota_1\circ f_1:U\to F_1\times F_2$. Similarly, $\iota_2\circ f_2:U\to F_1\times F_2$. Hence, their sum is also a function $U\to F_1\times F_2$. So, finally by composing with $B:F_1\times F_2\to G$, we get the function $g:U\to G$.
Now, $f_1,f_2$ are $C^r$ by assumption, $\iota_1,\iota_2$ are bounded and linear (which implies $C^{\infty}$ and hence trivially also $C^r$), so by the remark above, $\iota_1\circ f_1$ and $\iota_2\circ f_2$ are $C^r$ functions. Therefore, their sum is also $C^r$ (the fact that sum of $C^r$ functions is $C^r$ is an easy exercise in induction and the simple rule that $D(\alpha+\beta)_x= D\alpha_x+D\beta_x$). Finally $B$ is a bounded bilinear mapping, it is also $C^{\infty}$ hence $C^r$ (more generally, for any bounded $p$-multilinear mapping, if you take its $(p+1)^{th}$ derivative you get $0$... again an exercise in induction... therefore it is easily seen to be $C^{\infty}$). Therefore, $g=B\circ \left(\iota_1\circ f_1 + \iota_2\circ f_2\right)$ is the composition of $C^r$ functions hence $C^r$. In short,
\begin{align} g= B\circ \left(\iota_1\circ f_1 + \iota_2\circ f_2\right) ``=" C^{\infty}\circ \left(C^{\infty}\circ C^r + C^{\infty}\circ C^r\right) = C^r \end{align}
Edit:
Your calculation is right, the given formula is wrong, because the formula on the right is not symmetric in $h_1,\dots, h_k$. For example, when $k=2$, the given formula says that \begin{align}D^2g(x)(h_1,h_2) &= B(f_1(x), D^2f_2(x)(h_1,h_2)) + 2B(Df_1(x)(h_1), Df_2(x)(h_2)) \\ &+ B(D^2f_1(x)(h_1,h_2),f_2(x)). \end{align} The LHS $D^2g(x)$ being a second derivative must be symmetric in $h_1,h_2$, but the RHS is clearly not symmetric, so the given formula is wrong. If I'm not mistaken, the correct formula is obtained by symmetrizing your given formula.
So, if we let $\lambda_k(x):\underbrace{E\times\cdots \times E}_{\text{$k$ times}}\to G$ be the multilinear map \begin{align} &\lambda_k(x)(h_1,\dots, h_k)\\ &:= \sum_{j=0}^k \binom{k}{j} B\left (D^jf_1(x) (h_1,h_2,\cdots,h_j) ,D^{k-j}f_2(x) (h_{j+1},h_{j+2}, \cdots, h_k) \right ), \end{align} then $D^kg_x = \text{Sym}_k(\lambda_k(x))$, or more explicitly, \begin{align} D^kg(x)(h_1,\dots, h_k)&=\dfrac{1}{k!}\sum_{\sigma\in S_k}\lambda_{k}(x)\left(h_{\sigma(1)}, \dots, h_{\sigma(k)}\right), \end{align} where $S_k$ is the set of all permutations on $k$ objects (i.e the set of all bijections $\sigma:\{1,\dots, k\}\to \{1,\dots, k\}$).