$K_1\preccurlyeq K_2$ iff $C(K_1|_X,[0,1])\subseteq C(K_2|_X,[0,1])$ for any compactifications $K_1$ and $K_2$ What shown below is a reference from "Elementos de Topología General" by Fidel Cassarubias Segura and Ángel Tamariz Mascarúa

Achtung!!!
If $(h,K)$ is compactification of $X$ it is common to identify $X$ with $h[X]$, since $h$ is a homeomorphism between $X$ and $h[X]$. So using this identification we can consider $X$ as a dense subspace of any its compactification $K$. So using this agreement if $(h,K)$ is a compactification of $X$ and if $p:K\rightarrow Y$ is a function, then we use $p|_X$ to indicate $p|_{h[X]}$, that is $p\circ h$. Therefore we indicate with
  $$
C(K|_X,[0,1]):=\{g|_X:g\in C(K,[0,1])\}
$$
  the set of all restrictions to $X$ of the continuous functions from $K$ to $[0,1]$.
Lemma
Let be $K$ a hausdorff compactification of $X$ and let be 
  $$
F:=\Delta_{f\in C(K,[0,1])}f:K\rightarrow[0,1]^{C(K,[0,1])}
$$
  the diagonal function of the collection $C(K,[0,1])$. So
  $$
(F|_{X},F[K])
$$
  is a hausdorff compactification of $X$ that is equivalent to $K$
Achtung!!!
If $K$ is a hausdorff compactification  of $X$ then we define the function $r:C(K,[0,1])\rightarrow C(K|_X,[0,1])$ defined as
  $$
r(g)=g|_X
$$
  for any $g\in C(K,[0,1])$: so using the density of $X$ in $K$ it is possible to prove that $r$ is a bijection. Now if $I$ and $J$ are two index collections such that $I\subseteq J$ and if $\mathfrak{X}_I=\{(X_i,\mathcal{T}_i): i\in I\}$ and $\mathfrak{X}_J=\{(X_j,\mathcal{T}_j): j\in J\}$ are two collections of topological spaces for which there exist an injection $\phi:I\rightarrow J$ such that $(X_i,\mathcal{T}_i)=(X_{\phi(i)},\mathcal{T}_{\phi(i)})$ for any $i\in I$, we define the function 
  $$
\pi_I:\prod_{j\in J}X_j\rightarrow\prod_{i\in I}X_i
$$
  through the condition
  $$
\pi_I(f)=f|_I
$$
  for any $f\in\prod_{j\in J}X_j$: so using the universal mapping theorem for products it is possible to prove that $\pi_I$ is continuous. Then we name $\pi_I$ the projection of $\prod_{j\in J}X_j$ in $\prod_{i\in I}X_i$
Theorem
If $K_1$ and $K_2$ are two hausdorff compactifications of $X$ then it results that $K_1\preccurlyeq K_2$ if and only if it results that $C(K_1|_X[0,1])\subseteq C(K_2|_X,[0,1])$.
Proof. 
  
  
*
  
*We suppose that $K_1\preccurlyeq K_2$, thus there exist a continuous function $p:K_2\rightarrow K_1$ such that $p|_X=id_X$. So if $f\in C(K_1|_X,[0,1])$ then there exist $f^*\in C(K_1,[0,1])$ that extend $f$ to $K_1$. So it is easy to verify that 
  $$
g^*=f^*\circ p:K_2\rightarrow [0,1]
$$is a continuous extension of $f$ to $K_2$, thus we can claim that $f\in C(K_2|_X,[0,1])$.
  
*So now we suppose that $K_1$ and $K_2$ are two compactifications of $X$ such that $C(K_1|_X,[0,1])\subseteq C(K_2|_X,[0,1])$ and we prove that $K_1\preccurlyeq K_2$. So let be
  $$
F_1:=\Delta_{f\in C(K_1,[0,1])}f:K_1\rightarrow [0,1]^{C(K_1,[0,1])}
$$
  and
  $$
F_2:=\Delta_{g\in C(K_2,[0,1])}g:K_2\rightarrow[0,1]^{C(K_2,[0,1])}
$$
  the diagonal functions of $C(K_1,[0,1])$ and $C(K_2,[0,1])$. Then for the previous lemma we know that $(F_1|_X,F_1[K_1])$ and $(F_2|_X,F_2[K_2])$ are two compactifications equivalent to $K_1$ and $K_2$ and so to prove the statement it is sufficient to discover a continuous function $p:F_2[K_2]\rightarrow F_1[K_1]$ such that $p\circ(F_2|_X)=F_1|_X$. So since the functions
  $$
r_1:C(K_1,[0,1])\rightarrow C(K_1|_X,[0,1])
$$
  and
  $$
r_2:C(K_2,[0,1])\rightarrow C(K_2,[0,1])
$$
  are bijections, we can respectively suppose that $F_1$ and $F_2$ are embeddings of $K_1$ in $[0,1]^{C(K_1|_X,[0,1])}$ and of $K_2$ in $C(K_2,[0,1])$. Now let be $\pi_{C(K_1|_X,[0,1])}$ the projection of $[0,1]^{C(K_2|_X,[0,1])}$ in $[0,1]^{C(K_1|_X,[0,1])}$ and we define $p=\pi_{C(K_1|_X,[0,1])}|_{F_2[K_2]}$. So it results that 
  $$
(p(F_2(x)))(f)=(F_2(x)|_{C(K_1|_X,[0,1])})(f)=(F_2(x))(f)=f(x)=(F_1(x))(f)
$$
  for any $x\in X$ and $f\in C(K_2|_X,[0,1])$, that is what we wanted.
  

For the sake of completeness to follow I put the original text of the proof: I hope mine was a good translation.

So I desire to discuss the inclusion
$$
C(K_1|_X,[0,1])\subseteq C(K_2|_X,[0,1])
$$
that I don't understand completely: from the proof of the point 1. of the theorem it seems to me that it means that there exist an injective function $\phi$ from $C(K_1|_X,[0,1])$ to $C(K_2|_X,[0,1])$; is this could be true? Then it seems to me that at the end of the proof there is a typo that I corrected in the traslation: is my revision correct?
So could someone help me, please?
 A: The inclusion $C(K_1\restriction X, [0,1]) \subseteq C(K_1\restriction X, [0,1])$ can be seen literally, if we unwind the homeomorphisms.
Suppose we have $p: K_2 \to K_1$ with $p \circ h_2 = h_1$ from $(h_1, K_1) \preceq (h_2, K_2)$. If $g \in C(K_1\restriction X, [0,1])$ this means that $g : X \to [0,1]$ is continuous and there is a continuous $\overline{g}: K_1 \to [0,1]$ such that $\overline{g} \circ h_1 = g$; i.e. $\overline{g}$ is an "extension" of $g$ modulo $h_1$, as it were. But then we can define $\overline{g'}: K_2 \to [0,1]$ by $\overline{g} \circ p$ (also continuous) and then 
$$\overline{g'} \circ h_2 = (\overline{g} \circ p) \circ h_2 = \overline{g} \circ (p \circ h_2) = \overline{g} \circ h_1 = g$$ so that indeed (via the same convention/identification as before) $g \in C(K_2\restriction X, [0,1])$, so a literal inclusion, not just an injective map.
It seems to me, optically at least, that the diagram suggests we indeed need to consider the composition $p \circ (F_2\restriction X)$, but I have to check more carefully when I have more time tomorrow. It's a bit of a notational mess anyway, due to the extra homeomorphisms and hidden identifications. 
