Setup: A diagram
$\require{AMScd} \begin{CD} X @>>> Y\\ @VVV @VV{f}V\\ U @>>> V \end{CD}$
in a (proper) model category is called homotopy cartesian if there exists a factorisation $Y\xrightarrow{\sim}W\twoheadrightarrow V$ of $f$ such that there exists a weak equivalence $X\xrightarrow{\sim} W\times_VU$. While this definition makes sense, it seems that the naive way to define "cartesian up to homotopy" would be to just say that the diagram is homotopy cartesian if $X\xrightarrow{\sim} Y\times_VU$ without involving any factorisations.
Question 1: Why do we use the weaker condition?
Idea: Usually when the actual definition is weaker than the stronger (naive) one, it might be that the stronger definition isn't well behaved under certain operations. So I considered the property:
If in the diagram
$\begin{CD} X @>>> Y @>>> Y'\\ @VVV @VVV @VVV \\ U @>>> V @>>> V' \end{CD}$
both small squares are homotopy cartesian, then the outer square is also homotopy cartesian.
My guess was that this property works for the weaker definition but not for the stronger one. Indeed I found a proof which seems to rely on the fact that we may choose different factorisations. However it is obviously a different thing to say that my proof needs this property than to say any proof needs it. (The proof was an exercise so it might very well be that it is not the easiest one)
Question 2: Is my idea correct? Or is there a totally different answer to my first question?