Can every bilinear map be represented as $B(x, y) = A(x)y$? Let $B(x, y) : X \times Y \to Z$ be a bilinear map on finite-dimensional vector spaces. Is it true that every such map admits a unique representation $A(x)y$ where $A : X \to (Y \to Z)$ is a linear map and $A(x)$ is a linear operator?

As Ian has explained you, it is rather easy.
Here, I would like to give you the same "feeling of easiness" using the classical form of a bilinear map once a basis has been fixed, i.e.,
$$B(X,Y)=X^TAY \tag{1}$$
(see example below) where the vectors $x,y$ are assimilated to the column vectors $X,Y$ of their coordinates with respect to a fixed basis. Using (1), one can write:
$$B(X,Y)=X^TAY=X^TA^TY=(AX)^TY=(AX)\cdot Y \tag{2}$$
(assuming  matrix $A$ is symmetrical) where symbol $\cdot$ is for dot product.
(2) is nothing else that the coordinate form of your expression $A(X)(Y)$.
Example of a bilinear map under the coordinate form (1):
$$2x^2+10xy+3z^2=\underbrace{\begin{pmatrix}x&y\end{pmatrix}}_{X^T}\underbrace{\begin{pmatrix}2&5\\5&3\end{pmatrix}}_A\underbrace{\begin{pmatrix}x\\y\end{pmatrix}}_X.$$
