There is a canonical surjective $R$-algebra homomorphism $$\pi:R[x_1,\ldots,x_n]\to R[x_1,\ldots,x_n]/I$$ with $\ker\pi = I$, sending each polynomial $p(x_1,\ldots,x_n)$ to the coset $p(x_1,\ldots,x_n)+I$.
Suppose we have an $R$-algebra isomorphism $$g:R[x_1,\ldots,x_n]/I\stackrel{\sim}{\to} A.$$ Then the composition $f = g\circ\pi:R[x_1,\ldots,x_n]\to A$ is a surjective $R$-algebra homomorphism. Since $f$ is an $R$-algebra homomorphism,
$$f(p(x_1,\ldots,x_n)) = p(f(x_1),\ldots,f(x_n))$$ for each $p\in R[x_1,\ldots,x_n]$. It follows that since $f$ is surjective, every element of $A$ is a polynomial in $f(x_1),\ldots,f(x_n)$ with coefficients in $R$ -- that is, $A$ is generated by these $n$ elements as an $R$-algebra.
Let $A = B$ and $a_i = f(x_i)$. The first part says that there exists a unique $F$-homomorphism $\varphi \colon F[x_1,\dotsc,x_n] \to B$ such that $\varphi(x_i) = f(x_i)$ for each $i$.
Let $A = F[x_1,\dotsc,x_n]$ and $a_i = x_i$. The hypothesis in the second part says that there exists a unique $F$-homomorphism $\psi \colon B \to F[x_1,\dotsc,x_n]$ such that $\psi(f(x_i)) = x_i$ for each $i$.
Consider $\psi \circ \varphi \colon F[x_1,\dotsc,x_n] \to F[x_1,\dotsc,x_n]$. It is an $F$-homomorphism satisfying $\psi \circ \varphi (x_i) = x_i$. But, $\operatorname{id}_{F[x_1,\dotsc,x_n]}\colon F[x_1,\dotsc,x_n] \to F[x_1,\dotsc,x_n]$ is also an $F$-homomorphism satisfying $\operatorname{id}_{F[x_1,\dotsc,x_n]}(x_i) = x_i$. Taking $A = F[x_1,\dotsc,x_n]$ and $a_i = x_i$, the uniqueness condition in the first part says that $\operatorname{id}_{F[x_1,\dotsc,x_n]} = \psi \circ \varphi$.
Consider $\varphi \circ \psi \colon B \to B$. It is an $F$-homomorphism satisfying $\varphi \circ \psi (f(x_i)) = f(x_i)$. But, $\operatorname{id}_B \colon B \to B$ is also an $F$-homomorphism satisfying $\operatorname{id}_B(f(x_i)) = f(x_i)$. Taking $A = B$ and $a_i = f(x_i)$, the uniqueness condition in the hypotheses of the second part says that $\operatorname{id}_B = \varphi \circ \psi$.
Hence, $F[x_1,\dotsc,x_n] \cong B$.
Best Answer
First of all $\mathbb{C}[x_1 - a_1, \dots, x_n-a_n]=\mathbb{C}[x_1,\dots, x_n]$ and I think in the book you are reading it must be $$\phi : \mathbb{C}[x_1,\dots, x_n] \rightarrow \mathbb{C}[x_1, \dots, x_n], x_i \mapsto x_i - a_i.$$ This is easily seen an isomorphism. You also have two maps: $s_a : \mathbb{C}[x_1,\dots,x_n] \rightarrow \mathbb{C}, x_i \mapsto a_i$ whose kernel you want to determine and $s_0 : \mathbb{C}[x_1,\dots,x_n] \rightarrow \mathbb{C}, x_i \mapsto 0$ whose kernel you already know. Now check that $s_a\circ\phi=s_0$ and let's find out the kernel of $s_a$: $f\in\ker s_a$ iff $s_a(f)=0$ iff $s_0\circ\phi^{-1}(f)=0$ iff $\phi^{-1}(f)\in\ker s_0$ iff $f\in \phi(\ker s_0)$, so $$\ker s_a=\phi(\ker s_0).$$ Since we know that $\ker s_0=(X_1,\dots,X_n)$ it follows $$\ker s_a=\phi((X_1,\dots,X_n))=(\phi(X_1),\dots,\phi(X_n))=(X_1-a_1,\dots,X_n-a_n).$$