Showing that these two sets of linear transformations are same.

The sets $X$ and $Y$ are not the same, the elements of the first are linear maps $V\to W$ and those of the second are linear maps $V/U\to W$; elements of the quotient space $V/U$ are equivalence classes of the form $[v]:=v+U$ and there are $|V|/|U|$ such distinct elements, not the same as elements of $V$.

What you perhaps want to show is that $|X|=|Y|$ (ie, they have equal cardinality), for which you need to construct a bijection between the two. In fact, $X$ and $Y$ are isomorphic as algebraic structures when they are equipped with the binary operation of function addition, ie, $(X,+)\simeq (Y,+)$, a stronger property.

Consider the map $f\mapsto \Phi(f)$ mapping an element $f$ of $X$ (linear map $f\colon V\to W$ with $\ker f=U$) to the element $\Phi(f)$ of $Y$ (linear map $\Phi(f)\colon V/U\to W$ defined as $\Phi(f)(v+U)=f(v)$)

Show that the map $\Phi\colon X\to Y$ is well-defined, ie, if $f,g\in X$ and $f=g$, then $\Phi(f)=\Phi(g)$ (this might sound obvious but due to the abstract definition of $\Phi$, it is necessary to show that $\Phi$ is a well-defined, unambiguous map; in fact, this map is not well defined without the condition of $\ker f=U$)

Show that $\Phi$ is injective and surjective (the latter is obvious by construction but you should still explicitly write it out).

Show that $\Phi$ is a homomorphism under $+$, ie,

$$\bigl(\Phi(f+g)\bigr)(v+U)=\bigl(\Phi(f)+\Phi(g)\bigr)(v+U)$$

Note: The map addition $+$ is defined as $(f+g)(v):=f(v)+g(v)$ for elements $f,g\in X$ and as $(F+G)(v+U):=F(v+U)+G(v+V)$ for elements $F,G\in Y$