Consider $$\mathcal{B}(\mathbb{R}^n; \mathbb{R}^m) = \{ B: \mathbb{R}^n \times \mathbb{R}^n \rightarrow\mathbb{R}^m ; ~B \ \text{ is bilinear}\}$$
show that
\begin{split} \Phi: & \ \mathcal{L}\big( \mathbb{R}^n, \mathcal{L}(\mathbb{R}^n, \mathbb{R}^m) \big) \longrightarrow \qquad \mathcal{B}(\mathbb{R}^n; \mathbb{R}^m) \\ & \qquad \qquad T \qquad \ \ \quad \longmapsto \ \Phi(T)[x,y] := T(x)(y) \end{split}
is a canonical isomorphism.
My first question is: I must show that this is linear, right? So we do
\begin{split} \Phi(\lambda T+S)[x,y] & = (\lambda T+S)(x)(y) \\ & =(\lambda T(x)+S(x))(y) \\ & = \lambda T(x)(y) + S(x)(y) \\ & = \lambda \Phi(T)[x,y] +\Phi(S)[x,y]. \end{split}
I think that it's right, but i would appreciate any comment on this. Moving on, we know that $$Dim \ \mathcal{L}\big( \mathbb{R}^n, \mathcal{L}(\mathbb{R}^n, \mathbb{R}^m) \big) = Dim \ \mathcal{B}(\mathbb{R}^n; \mathbb{R}^m)$$
so it is enought to prove that $\Phi$ is injective, I'm working on that part, but i would like to know if it is the "right" idea.
Also, I'm having some hard time on understand how the elements of $\mathcal{L}\big( \mathbb{R}^n, \mathcal{L}(\mathbb{R}^n, \mathbb{R}^m) \big).$ Any recomendation on references in this topic will help me!
Your proof that $\Phi$ is linear is ok, but before moving on, although trivial, you should at least mention that $\Phi$ does take values in $\mathcal{B}(\Bbb R^n;\Bbb R^m)$. The quantity $\Phi(T)(x,y)=T(x)(y)$ is bilinear in $x$ and $y$ because $T$ is linear and because $T(x)$ is linear.
That said, given $B\in \mathcal{B}(\Bbb R^n;\Bbb R^m)$, let $\Psi(B):\Bbb R^n \to \mathcal{L}(\Bbb R^n;\Bbb R^n)$ be given by $\Psi(B)(x)(y)=B(x,y)$. Show that $\Psi$ is the desired inverse.