At page 19 in this book $f:X\to Y$ is defined to be $$f(a):=(\tilde\varphi(T_1')(a),\dots,\tilde\varphi(T_n')(a)).$$
To explain the notation above, $X\subseteq \mathbb{A}^m(k)$, $Y\subseteq \mathbb{A}^n(k)$ are affine algebraic sets, and $\tilde\varphi:k[T_1',\dots,T_n']\to k[T_1,\dots,T_m]$ is a $k$-algebra morphism of polynomial rings that induces a morphism $\varphi:k[T_1,\dots,T_m]/I(Y) \rightarrow k[T_1',\dots,T_n']/I(X)$. Here, $I(X)$ and $I(Y)$ are the ideals of polynomials vanishing on $X$ and $Y$, respectively.
My question is: How do we know that $f(a)\in Y$ for all $a \in X$?
First, note that in order for $\tilde\varphi$ to induce a morphism of affine coordinate rings as you describe, we must have that
$$g \in I(Y) \implies \tilde\varphi(g) \in I(X). \,\,\, (*)$$
The map $\tilde\varphi$ is determined by $T_i' \mapsto g_i(T_1,\dots,T_m)$, for some $g_i \in k[T_1,\dots,T_m]$. We can consider the map $f:X \rightarrow \mathbb A^n(k)$ defined by $a \mapsto (g_1(a),\dots, g_n(a))$, and the problem is to verify that $f(X) \subset Y$. For this, it suffices to show that $h\in I(Y), a \in X \implies h(f(a)) =0,$ since $Y$ is the subset of $\mathbb A^n$ where all polynomials in $I(Y)$ vanish.
Observe that $h(f(a))=h(g_1(a),\dots,g_n(a))=\tilde\varphi(h)(a)$. Hence, for $h \in I(Y)$ and $a \in X$, $(*)$ implies that $h(f(a))=0.$