I actually think something like the quasi-compactness of $Z$ is necessary (or at least, I haven't been clever enough to figure out how to do it otherwise), but not really for the reasons why Justin asked the question. But the hypotheses on $f,g$ seem unnecessary…
Stephen's answer is completely right if we assume the morphisms $f \colon X \to Y$ and $g \colon Y \to Z$ are projective, i.e., $i \colon X \hookrightarrow \mathbb{P}^m_Z$ and $j \colon Y \hookrightarrow \mathbb{P}^n_Z$ are closed immersions, but according to Hartshorne, a sheaf is very ample if it is induced by the pullback of $\mathcal{O}(1)$ through an arbitrary immersion, defined as the composition of an open immersion followed by a closed immersion.
This has the consequence that the composition of two immersions may not be an immersion since the composition of a closed immersion followed by an open immersion is not necessarily an immersion in Hartshorne's sense (see [Stacks, Tag 01QW]—note the difference in terminology). So, the composition $(1 \times \sigma) \circ (j \times 1) \circ i$ in Stephen's answer might not be an immersion!
There are a few ways to make this exercise solvable, disregarding the fact that Hartshorne's definition of very ample should probably be fixed:
a) Redefine an immersion to be what is in EGA—that is, an open immersion followed by a closed immersion, what I will call a "locally closed" immersion [EGAI, 4.1.3 and 4.2.1]. In this case, the composition of two locally closed immersions is again a locally closed immersion by [EGAI, 4.2.5], and so Stephen's argument goes through. In particular, it seems the assumptions on $f$ and $g$ are unnecessary for the statement of the problem with Hartshorne's definition of very ample.
b) Assume that $j \colon Y \hookrightarrow \mathbb{P}^n_W$ is quasi-compact. Factor $i \colon X \hookrightarrow \mathbb{P}^m_Z$ as $i = i_2 \circ i_1$, where $i_2$ is a closed immersion and $i_1$ is an open immersion.
Then, $(j \times 1)$ is quasi-compact, hence $(j \times 1) \circ i_2$ is quasi-compact, and so by [Stacks, Tag 01QV], $(j \times 1) \circ i_2$ is an immersion hence $(1 \times \sigma) \circ (j \times 1) \circ i_2 \circ i_1 = (1 \times \sigma) \circ (j \times 1) \circ i$ is an immersion.
c) Assume $Z$ is locally noetherian (which jives well with Hartshorne's use of noetherian hypotheses—see p. 100). Then, $\mathbb{P}_Z^{mn+m+n}$ is locally noetherian, hence by [Stacks, Tag 01OX] a locally closed immersion into $\mathbb{P}_Z^{mn+m+n}$ is quasi-compact, so $(1 \times \sigma) \circ (j \times 1) \circ i_2$ is again quasi-compact and we are in situation b).
Now the question becomes how these assumptions relate to those in [EGAII, 4.4.10(ii)] you mentioned. Does $g$ being quasi-compact and $Z$ being quasi-compact imply $j$ is quasi-compact, so we are situation b)? I don't know…
1) comes down to a statement from multilinear algebra: if $V$ is a free $k$-module of rank $n$, there is a natural pairing $\wedge^n(V) \otimes \wedge^n(V^*) \to k$ given by mapping $v_1 \wedge \dotsc \wedge v_n \otimes \phi_1 \wedge \dotsc \wedge \phi_n$ to the determinant of the matrix $(\phi_i(v_j))$. It is perfect, i.e. induces a natural isomorphism $\wedge^n(V)^* \cong \wedge^n(V^*)$. You can check this directly using bases.
For the rest of the questions, you should explaind your notation ($Y,X,I$, etc.).
Best Answer
In this case, Hartshorne is being sloppy and using $\mathcal{I}$ and $\mathcal{I}_Y$ interchangably for the sheaf of ideals defining $Y$ as a closed subscheme of $X$ (see Proposition II.8.12 and Theorem II.8.17 for a reminder about what $\mathcal{I}/\mathcal{I}^2$ means here). As for an explanation of the isomorphism, recall that $\mathcal{O}_Y=\mathcal{O}_X/\mathcal{I}_Y$, so the tensor product is just $\mathcal{I}_Y/\mathcal{I}_Y^2$ by the standard natural isomorphism $M\otimes_R R/I = M/IM$.