Show that if $T$ is a Skolem theory in $L$, then $T$ is a Skolem theory in $L'$

Let $$L$$ be a first order language and $$L'$$ a language which comes from $$L$$ by adding constants. Show that if $$T$$ is a Skolem theory in $$L$$, then $$T$$ is a Skolem theory in $$L'$$ (and hence so is any theory $$T\subseteq T'$$ in $$L'$$).

This is question 3.1.1 from Hodges' Shorter Model Theory; I found a solution (as attached below) but I don't at all understand what he is doing. Specifically:

1. What is he doing when he 'write $$\phi$$ as $$\phi '(\bar{x},\bar{c},y)$$, with $$\phi' (\bar{x},\bar{z},y)$$'? Supposedly $$\phi$$ is expanded from having only 2 variables ($$\bar x, y)$$ to 3 ($$\bar x,\bar c,y$$) because of the language expansion by constants...? But why then replace the constant $$\bar c$$ with $$\bar z$$, a new free variable?

2. What is this lemma on constant he is citing? I have been looking all over 3.1 but I don't see any thing related. Without knowing what this lemma is, I can't understand what that step of the proof is about.

The point of replacing $$\phi(\bar{x}, y)$$ by $$\phi'(\bar{x}, \bar{z}, y)$$, is to get an $$L$$-formula (which is used in the next line of the proof). The formula $$\phi(\bar{x}, y)$$ is an $$L'$$-formula, and since we have only added constants in $$L'$$, the only symbols that prevent $$\phi(\bar{x}, y)$$ from being an $$L$$-formula are those constants. Another way of formulating this is that every $$L'$$-formula $$\phi(\bar{x}, y)$$ arises from some $$L$$-formula $$\phi'(\bar{x}, \bar{z}, y)$$ by filling in some tuple of constants $$\bar{c}$$ on the place of $$\bar{z}$$. So $$\phi(\bar{x}, y)$$ is $$\phi'(\bar{x}, \bar{c}, y)$$.
For any theory $$T$$ and any formula $$\phi(\bar{x})$$, both not containing the constants $$\bar{c}$$, we have: $$T \vdash \forall \bar{x} \phi(\bar{x}) \quad \Longleftrightarrow \quad T \vdash \phi(\bar{c}).$$
The implication from the left to the right is trivial (and even holds if $$\phi$$ or $$T$$ contains $$\bar{c}$$). The other direction is not needed for your question, but may be a nice exercise to prove. So a direct application of this lemma (left to right direction) allows you to replace $$\bar{z}$$ by $$\bar{c}$$ in that step.