Theorem: There exists an isomorphism between the space of bilinear forms $\varphi$ on vector space $L$ and space $\mathcal{L}(L,L^*)$ of linear transformations $\mathcal{A}:L\mapsto L^*$.
I'm struggling understanding the proof of it, particularly the first part, where it's shown that $\mathcal{A}$ is linear function.
Proof (first part)
Let $\varphi(x,y)$ be a bilinear form on $L$. Now we'll associate this bilinear form with transformation $\mathcal{A}:L\mapsto L^*$ in the following way: $y$ is assigned to $\mathcal{A}$ so that $\mathcal{A}(y)$, i.e $\mathcal{A}:y\mapsto \mathcal{A(y)}$. To vector $x$ we'll assign $\mathcal{A}_x$, i.e $x\mapsto\mathcal{A}_x$ and $\varphi(x,y)=(x,A_x(y))$ and let those two numbers be the same. Since $\varphi(x,y)=(x,\mathcal{A}_x(y))$ and due to the fact that $\varphi(x,y)$ is linear by both arguments we conclude that $\mathcal{A_x(y)}$ must be a linear function.
Confusion with the proof
Knowing the fact that $\varphi$ is bilinear function on $L$ it should be that $\varphi:L\times L\mapsto \mathbb{F}$, where $\mathbb{b}$ is field of scalars over $L$. But if we look at what was written in the proof, $\varphi(x,y)=(x,\mathcal{A}_x(y))$, wouldn't this mean that $\varphi:L\times L\mapsto L\times L^*$? How is $(x,\mathcal{A}_x(y))$ supposed to be a scalar? And then after that is defined comes the next part of the sentence "and let those two numbers be the same". What does that mean?
Maybe I might just be having a flawed understanding of space $\mathcal{L}(L,L^*)$ and its linear transformations $\mathcal{A}$. Anyways, any help will be appreciated.
$\newcommand{\A}{\mathcal{A}}$The proof makes absolutely no sense at the start.
"$\A$ assigns $y$ with $\A=y$" - meaningless. "$\A:y\to\A(y)$" - incorrect notation (should be $y\mapsto\A(y)$) but also, still meaningless, since $\A(y)$ is still an undefined symbol. $x\mapsto\A_x$ still makes no sense because $\A_x$ is still undefined. They are using their symbols before defining them, and then quite sloppily defining them after the fact. I don't you blame for being confused, because that was pretty badly written.
In this post I ignore continuity considerations (usually $L^\ast$ denotes continuous linear functionals, I assume you just mean, linear functionals in general - which I have seen written as $L^\sharp$).
Let's give it another go. Fix a bilinear $\varphi:L\times L\to\Bbb F$. For any fixed $x\in L$, the map $L\ni y\mapsto\varphi(x,y)\in\Bbb F$ is linear, so you have an associated element of $L^\ast$. Therefore, $\A:L\to L^\ast,\,x\mapsto(y\mapsto\varphi(x,y))$ is a well defined function.
Is it linear? Yes. For scalars $\alpha,\beta\in\Bbb F$ and any $x,x'\in L$, we have that $\A(\alpha x+\beta x')$ is the functional: $$y\mapsto\varphi(\alpha x+\beta x',y)=\alpha\varphi(x,y)+\beta\varphi(x',y)=\alpha\A(x)(y)+\beta\A(x')(y)$$So it follows that $\A(\alpha x+\beta x')=\alpha\A(x)+\beta\A(x')$ - $\A$ is linear.
This defines a map $\Gamma:\mathrm{Bil}(L,L;\Bbb F)\to\mathscr{L}(L,L^\ast)$. In fact, if you equip both spaces with their natural linear structures, you can check that $\Gamma$ is itself linear.
In case the rest of the proof is confusing, I will define an inverse to $\Gamma$ for you. Given a linear $\A:L\to L^\ast$, we can define a map: $$\varphi:L\times L\ni(x,y)\mapsto\A(x)(y)\in\Bbb F$$Fix $\alpha,\beta\in\Bbb F$ and $x,x',y,y'\in L$.
$$\begin{align}\varphi(\alpha x+\beta x',y)&=\A(\alpha x+\beta x')(y)\\&=(\alpha\A(x)+\beta\A(x'))(y)\\&=\alpha\A(x)(y)+\beta\A(x')(y)\\&=\alpha\varphi(x,y)+\beta\varphi(x',y)\end{align}$$
So $\varphi$ is linear in the first argument. I invite you to check $\varphi$ is also bilinear. It is then hopefully clear that we get a (linear) inverse map to $\Gamma$ that runs $\mathscr{L}(L,L^\ast)\to\mathrm{Bil}(L,L;\Bbb F)$.