I'm reading some lecture notes on homotopy, and the author has just proved the theorem:
If $f_{0} \simeq f_{1}$ and $g_{0} \simeq g_{1}$, then $g_{0} \circ f_{0} \simeq g_{1} \circ f_{1}$
Where, for each $i \in \{0,1\}$, $f_{i} : X \to Y$ and $g_{i}: Y \to Z$, and $\simeq$ denotes homotopy equivalence.
He does this by showing $g_{0} \circ f_{0} \simeq g_{0} \circ f_{1} \simeq g_{1} \circ f_{1}$
For the first equivalence he uses the homotopy $g_{0} \circ F$ (where $F$ is a homotopy from $f_{0}$ to $f_{1}$) and for the second he uses $G \circ (f_{1} \times Id_{[0,1]})$. I completely understand that. But I don't see why you couldn't (or wouldn't) use the quicker:
$$H(x,t)=G(F(x,t),t)$$
This is continuous (unless there's something key I've misunderstood), and $H(x,0)=G(F(x,0),0)=g_{0}(F(x,0))=g_{0}(f_{0}(x))$ and similarly for $t=1$.
Best Answer
There's no strong reason not to use $G(F(x,t),t)$. One reason the author may not have is that they may have had in mind the general principle that you can concatenate homotopies, and that this often lets you build homotopies that "do multiple things" by doing each one separately and concatenating the pieces along the time coordinate. In this particular example, it is actually possible to do both of them at once as you have pointed out, but it isn't always (see for instance my answer to this question).