If we have a homotopy pullback square
$$ \begin{matrix}A &\longrightarrow{}& Y \\ \ \ \downarrow & & \ \ \downarrow \\ X &\longrightarrow{}& Z \end{matrix} $$
Question:
How can we show that there is a fiber sequence $$ \Omega Z\to A \to X\times Y\to Z $$ where $\Omega Z$ is the loop space of $Z$. Is my statement true in general?
Do we have $$ ... \to \Omega^2 Z \to \Omega A\to \Omega ( X\times Y) \to \Omega Z\to A \to X\times Y\to Z, $$ in general?
Thanks for the answer. (Ref is welcome!)
Firstly, I don't believe that the first statement is true without additional assumptions. There is no natural way to obtain a map $X\times Y\rightarrow Z$ from the given data in general. If $Z$ is an $H$-space (in particular if it is a loop space) then the consution is possible. Likewise, if of one of the maps $X\rightarrow Z$, $Y\rightarrow Z$ has a weakly principal homotopy action on $Z$, then the construction should go through. Anyway, I'll focus on producing the fibration sequence terminating
$$\dots\rightarrow\Omega X\times\Omega Y\xrightarrow{\Delta}\Omega Z\rightarrow A\xrightarrow{\varphi} X\times Y,$$
which does exist for any homotopy pullback square as in your question.
Let $(E)$ be the homotopy pullback diagram
$\require{AMScd}$ \begin{CD} A @>g>> Y\\ @VfV V @VkVV\\ X @>h>> Z. \end{CD}
Now, since every object in $Top$ is fibrant, the square
$\require{AMScd}$ \begin{CD} X\times Y @>pr_2>> Y\\ @Vpr_1V V @VVV\\ X @>>> \ast. \end{CD}
is both a topological and homotopy pullback. Call this square $(B)$. Then the identities on $X$ and $Y$, and the unique map $Z\rightarrow \ast$ induce a morphism of pre-pullback data
$$(X\xrightarrow{h} Z\xleftarrow{k} Y)\Rightarrow (X\rightarrow\ast\leftarrow Y)$$
which thus gives a morphism of homotopy pullback squares $(E)\Rightarrow (B)$, and in particular an induced map
$$\varphi:A\rightarrow X\times Y$$
by the weakly universal property of the homotopy pullback squares. This map $\varphi $ is not unique, but since it must satisfy
$$pr_1\circ\varphi\simeq f,\qquad pr_2\circ\varphi\simeq g,$$
and any (homotopy class of) map into a product is determined by its projections, we see that
$$\varphi\simeq (f,g).$$
Now the data of the morphism $(E)\Rightarrow (B)$ displays as a homotopy commutative cube (which I am in no way attempting to draw in Mathjax). If you take the homotopy fibes of the vertical maps, then you get another cube, which we'll conveniently call $(F)$,
$\require{AMScd}$ \begin{CD} F_{\varphi} @>>> \ast\\ @VV V @VVV\\ \ast @>>> Z. \end{CD}
Here the bottom right-hand space is the homotopy fibre of $Z\rightarrow Z$, which is equivalent to $Z$. The top right-hand space is the homotopy fibre of $id_Y:Y\rightarrow Y$, which is contracible. Similarly for the bottom left-hand space. The top space $F_\varphi$ is the homotopy fibre of $\varphi=(f,g):A\rightarrow X\times Y$. All the maps in the square are induced as maps between homotopy fibres in the standard way from homotopy commuative diagrams of fibrations.
Now it is a standard theorem that the square of homotopy fibres of a morphism of homotopy pullback diagrams is again a homotopy pullback diagram. In particular this tells us that the square $(F)$ is a homotopy pullback diagram, and it is a simple exercise to verify that the homotopy pullback of $(\ast\rightarrow Z\leftarrow \ast)$ (the space $Z$ is assumed to be pointed) has the homotopy type of the loop space $\Omega Z$. Hence
$$F_\varphi\simeq \Omega Z.$$
Thus, choosing suitable maps we get the homotopy fibration claimed to exist at the beginning of this answer.
Now, as a bonus round, I've labelled the fibration connecting map $\Delta:\Omega X\times\Omega Y\rightarrow \Omega Z$. If you do the hard work of writing everything down explicitly, and choose your maps and homotopies wisely, then it is possible to show that
$$\Delta\simeq \Omega h\circ pr_1-\Omega k\circ pr_2,$$
where the maps are subtracted using the loop adition on $\Omega Z$. That the map should have this form is intuitively clear, since we must have $\Delta\circ\Omega\varphi\simeq\ast$, and this certainly holds with $\varphi\simeq (f,g)$.
The point is that this is the reason why your question $1)$ was not precise. Without the loop addition there is no way to combine the maps $h,k$ to get a map from $X\times Y$. If $Z$ is an $H$-space, then you might like to sort out my claims on extending the sequence as an exercise.
As for you question $2)$, well the comments still hold for the map on the right-hand side. And there appears to be a typo (or lack of bracketing) in the middle. But why would this not exist given that it is just the continuation of an existing homotopy fibration sequence.