I'm following the discussion from here.
If you give me a map $U \stackrel{f}{\to} V$ from one test space to another, then for any element $V \stackrel{p}{\to} X$ of $X(V)$, I can produce an element $U \stackrel{f}{\to} V \stackrel{p}{\to} X$ of $X(U)$. In other words, your map $U \to V$ induces a transformation $X(U) \leftarrow X(V)$.
I can't get, why it induces map $X(U) \leftarrow X(V)$ and not $X(U) \rightarrow X(V)$ ?
I thought that this map would produce functorial diagram like that
$$ \require{AMScd} \begin{CD} U @>{f}>> V;\\ @VVV @VVV \\ X(U) @>{X(f)}>> X(V); \end{CD} $$ but that seems not to be the case. What's the idea behind reverse morphism arrows?
Algebraically, your quote already tells you what's going on: if you start with $V \to X$, then you can produce $U \to X$. Therefore you get a function $\hom(V, X) \to \hom(U,X)$.
The 'reference' example here is the notion of restricting a function. For example, let $C(X)$ be the (contravariant) functor of continuous, real-valued functions on $X$.
If $f$ is a continuous function on $X$ and $S \to X$ is the inclusion of a subspace, then you can form the restriction $f|_S$, which is a continuous function on $S$. This defines an operation $C(X) \to C(S)$.
(in fact, this operation is the value of $C$ at the inclusion arrow)