Tldr; jump to the $(\ast)$.
The category $Top$ of topological spaces is complete and cocomplete. That is, all limits and colimits exist in the category. On the other hand, the homotopy category $hTop$, formed by quotienting out the (weak) equivalences, is neither complete nor cocomplete. In fact while it has products and coproducts, it has very few other limits or colimits. See here for some examples of pullback/pushout squares which do not exit in $hTop$.
The point is that the strict fibre of an arbitrary map $f:X\rightarrow Y$ over a point $y\in Y$ is the equaliser of the two maps $f,c_y:X\rightrightarrows Y$, where $c_y$ is the constant map at $y$. Now this is not a good homotopical notion, in general, due to the lack of limits in $hTop$. It is a theorem that if a category has equalizers and finite products then it has all finite limits, and we have already noted above that $hTop$ does not even have all pullbacks. Since it has products $hTop$ cannot have all equalisers.
To see exactly what goes wrong consider the fact that the strict fibres of $f:X\rightarrow Y$ over the different points of $Y$ will not in general have the same homotopy type (even keeping within the same path component). For instance consider the map $S^1\rightarrow \mathbb{R}$ induced by projecting onto the first coordinate; the fibres are either empty, have one point, or have two points.
A sufficient condition that all the strict fibres over all points in a given path component have the same homotopy type is that the map $f$ is a fibration. And this is what the homotopy fibre does: it replaces the map $f$ by a fibration $p_f:E(f)\rightarrow Y$ and then takes its strict fibre over a given point. For everything to make sense we require some comparison map $j_f:X\rightarrow E(f)$ which is a homotopy, or at least weak, equivalence, and that everything should be natural - in the mathematical sense - at least up to homotopy.
Note that this 'up-to-homotopy' comparison map $j_f$ is the best we can do, for if $j_f$ were a homeomorphism, then the map $f$ would already be a fibration, and its homotopy fibres would just be its strict fibres. Thus if it were always possible to replace a map by a fibration up to homeomorphism, then there would be no need for homotopy fibres to begin with.
$(\ast)$ Now, the point is that the homotopy fibre of a map $f:X\rightarrow Y$ does not really live in $Top$. Since at best we can ask for the map $j_f$ to be a homotopy equivalence, the homotopy fibre of $f$ really lives in the homotopy category $hTop$ where it makes sense to consider the objects $X$ and $E(f)$ as the same. And now in the homotopy category everything is perfectly canonical, since $X$ and $E(f)$ are the same object as seen through categorical eyes. For example in the path space fibration we no longer have to pick an evaluation, since all these maps are homotopy equivalent, and thus represent the same morphism in $hTop$.
Thus we realise that the object $E(f)$ is nothing but an auxiliary construct. It is a point set lift of the problem in $hTop$ to a problem in $Top$. It allows us to create a more understandable model for the problem by using spaces with points, and maps defined on elements, rather than having everything 'up-to-homotopy'. However there is no canonical way to achieve this. For, just as the strict fibres of an arbitrary map $f:X\rightarrow Y$ may be badly behaved, the fibres of the projection $Top\rightarrow hTop$ are not uniform. Asking for a canonical section of this map is too much.
Thus we content ourselves with mathematically natural (again, up to homotopy) constructions point-set constructions, happy with the knowledge that once we project the problem back to $hTop$ we truly do have something safe and canonical.
Best Answer
I'm going to expand on my comments and given a more complete answer.
So far we have that $\newcommand\hofib{\operatorname{hofib}}\hofib(f)\to X$ is a Hurewicz fibration, because it's the pullback of a Hurewicz fibration. Moreover its fiber $F$ is the same as the fiber of $PY\to Y$. This is a general fact that follows from pullback pasting.
Consider the following diagram where $x\in X$ is a point, and both squares are pullbacks. $$ \require{AMScd} \begin{CD} F @>>> f^*P @>>> P\\ @VVV @VVV @VVV\\ x @>>> X @>>> Y \end{CD} $$ In this case $F$ is the fiber of $f^*P$ over $x$. However the outer square is also a pullback, by pullback pasting, so $F$ is also the fiber of $P$ over $f(x)$.
So the fiber of $\hofib(f)\to X$ is $\Omega Y$. All we need to do now is show that the fiber of a Hurewicz fibration is weakly equivalent to its homotopy fiber, where the homotopy fiber is defined by pulling back the path space of the codomain to the domain.
Let's assume then that $f:X\to Y$ is some general Hurewicz fibration (of pointed spaces). Let $F=f^{-1}(*)$ be the point-set fiber of $f$. Let $G = \hofib(f) = f^* PY$. We want to produce a homotopy equivalence between $F$ and $G$. For $PY$ I'm going to assume that we take paths that end at $*$ to fix a convention. Concretely then as a set, $G$ is the set of pairs $(x, \alpha)$ where $x\in X$ and $\alpha : I\to Y$ satisfies $\alpha(0)=f(x)$ and $\alpha(1)=*$.
First of all, we always have a map $i:F\to G$, given by $x\mapsto (x,*)$, where $*\in PY$ denotes the constant path at the base point. To get a map $G\to F$ we need to use the fact that $f$ is a Hurewicz fibration. Consider the square $$ \begin{CD} G @>\pi_1>> X \\ @VG\times \{0\}VV @VVfV \\ G\times I @>>\operatorname{ev}> Y, \end{CD} $$ where the bottom map is $(x,\alpha,t)\mapsto \alpha(t)$. The fact that this square commutes is the fact that if $(x,\alpha)\in G$, then $f(x) = \alpha(0)$.
Thus there exists a lift $H:G\times I \to X$. $H_0=\pi_1$, and $f\circ H_1 = *$, since $\alpha(1)=*$ for all $(x,\alpha)\in G$. Thus $H_1$ actually lands in the fiber, $F$. For convenience let's let $h=H_1$. Let's keep in mind for later use that $H$ is a homotopy between $\pi_1$ and $h$.
We now have $i:F\to G$ and $h:G\to F$. We just need to check that these maps are homotopy inverses of each other. Now $h\circ i$ is homotopic to $\pi_1\circ i$, but $\pi_1\circ i = 1_F$ by definition of $i$. For the other composition $i\circ h$, consider some element $(x,\alpha)\in G$. $h(x,\alpha) = x'$ for some $x'\in F$, and $H(x,\alpha,-)$ is a path from $x$ to $x'$ in $X$ lifting $\alpha$. This gives a path in $G$ from $(x', *)$ to $(x,\alpha)$ by $t\mapsto (H(x,\alpha,t), s\mapsto \alpha(t+s)$ (extend $\alpha$ so that $\alpha(t+s)=*$ when $t+s>1$). But this is a homotopy from $1_G$ to $i\circ h$: $$ (g,t)\mapsto \left(H(g,t), s\mapsto \operatorname{ev}(g,t+s)\right), $$ as desired.