If $i : A \to X$ and $j : X \to Z$ are cofibrations prove that the composition $j \circ i : A \to Z$ is a cofibration.
Since $i : A \to X$ is a cofibration for any space $Y$, a homotopy $F: A \times I \to Y$ and a map $f :X \to Y$ there exists $\tilde{F}:X \times I \to Y$ such that $$\tilde{F}(i(a),t)=F(a,t).$$
Similarly since $j :X \to Z$ is a cofibration for $G: X\times I \to Y$ and $g : X \to Y$ there exists an extension $\tilde{G}:Z\times I \to Y$ such that $$\tilde{G}(j(x),t)=G(x,t).$$
Now let $Y$ be any space and $H :A \times I \to Y$ a homotopy and $h : Z \to Y$ a map. I need to construct $\tilde{H}:Z \times I \to Y$ such that $$\tilde{H}(j(i(a)),t)=H(a,t).$$
Can I define $$\tilde{H}(z,t)=\begin{cases}\tilde{F}(z,t), & z\in X \\ \tilde{G}(z,t), & z\in Z\end{cases}?$$ This would satisfy the wanted condition.
You can't define it using $\tilde{F}$ because $\tilde{F}$ has not been defined (what is $F$? What is $f$? You need to provide these!).
Hint: restrict $h$ so that you do get suitable "$F$,$f$" and a suitable $\tilde{F}$ using $i$. Then do it all over again using $j$. Lift one step at a time.