A preliminary lemma:
Lemma: Let $I\subset k[x_1,\cdots,x_n]$ be an ideal and let $f\in k[x_1,\cdots,x_n]$ be a nonzero element. The closure of $V(I)\cap D(f)$ is given by the vanishing locus of the ideal $$J=\left(\frac{e}{f^{\deg_f e}}\mid e\in I\right),$$ where $\deg_f e$ is the number of times $f$ divides $e$.
Proof: The morphism of varieties $V(I)\cap D(f) \to \Bbb A^n$ corresponds to the ring map $k[x_1,\cdots,x_n]\to k[x_1,\cdots,x_n]_f/I_f$, and the closure of $V(I)\cap D(f)$ is the variety cut out by the kernel of this ring morphism. As $I_f=\left\{ \frac{e}{f^m}\mid e\in I, m\in\Bbb Z_{\geq0} \right\}$, if $e\in I$ then if $f^m\mid e$, we have $\frac{e}{f^m}$ is an element in $k[x_1,\cdots,x_n]$ which maps to zero. As every such element is $f^l\cdot \frac{e}{f^{\deg_f e}}$ for some $l\geq 0$, we have the result. $\blacksquare$
On to the exercise.
By remark 9.11, we know that $\widetilde{X}$ is the closure of $\pi^{-1}(X\setminus\{0\})$ in $\widetilde{\Bbb A^n}$. On $U_1$, we have that $\pi^{-1}(X)$ is the variety cut out by the ideal $(f(x_1,x_1y_2,\cdots,x_1y_n)\mid f\in I)$, while $\pi^{-1}(0)=V(x_1)$. As $\pi^{-1}(X\setminus \{0\})$ is $\pi^{-1}(X)$ without $\pi^{-1}(0)$, we see that on $U_1$, the closure of $\pi^{-1}(X\setminus \{0\})$ is exactly the desired ideal by our lemma. This proves (a).
For (b), we use the computation from (a). The exceptional divisor is the intersection of $\widetilde{X}$ with the $\Bbb P^{n-1}$ living over the origin in $\widetilde{A^n}$, so we can find this on $U_1$ by setting $x_1=0$. As evaluating $\frac{f(x_1,x_1y_2,\cdots,x_1y_n)}{x_1^{\deg_{x_1} f}}$ at $x_1=0$ gives the initial term of $f(y_1,\cdots,y_n)$ and then sets $y_1=1$, we see that the exceptional set is cut out by the ideal generated by $f^{in}(y_1,\cdots,y_n)$ as $f$ runs over the polynomials in $I$.
In order to attack (c), expand $f=f_r+f_{r+1}+\cdots$ and $g=g_s+g_{s+1}+\cdots$ in to homogeneous parts: then $fg=f_rg_s+(f_rg_{s+1}+f_{r+1}g_s)+\cdots$, and thus $(fg)^{in}=f^{in}\cdot g^{in}$. If $I=(f)$, this shows that the ideal we constructed in (b) is just $(f^{in})$. The counterexample is a standard sort of trick: consider the ideal $(x,x-y^2)\subset k[x,y,z]$. Then the initial ideal of this is $(x,y^2)$, but using the set of generators $\{x,x-y^2\}$, the ideal which is generated by the initial terms is just $(x)$.
The points $(p,[a:b]) \in \mathbb A^1 \times \mathbb P^1$ in the exceptional divisor parametrize pairs of a point $p$ on the center of the blow up (the line $Z = V(x_1,x_2)$) and a normal direction to $Z$ at $p$, i.e. a point of $\mathbb P(T_p\mathbb A^3/T_p Z)\cong \mathbb P^1$.
The connection to the case of blowing up a point in a variety $p \in X$ is that "the tangent space to a point" is trivial, so the normal space at $p$ is just the tangent space $T_p X$ without modding out by any subspaces.
In general (at least for blowing up a smooth center $Z$ in a smooth variety $X$), the exceptional divisor is isomorphic to the projectivization of the normal bundle $N_{Z\subset X}$.
Best Answer
Lemma 9.15 of Gathmann shows that the blow-up of $X$ at some functions $f_1, ..., f_r\in A(X)$ depends only on the ideal $(f_1, ..., f_r)\unlhd A(X)$.
This shows that if we have an isomorphism $f: X\to Y$, and if the ideal $J\unlhd A(Y)$ is sent to $I\unlhd A(X)$ under the induced isomorphism $f^*: A(Y)\to A(X)$, then the blow-up of $X$ at $I$ will be isomorphic to the blow-up of $Y$ at $J$.
We can show with the Affine Jacobi Criterion (Prop. 10.11) that the only singular points are $0$ in both $X_k$ and $X_l$. In our case this is equivalent to the tangent space having dimension at least 2 (since the local dimension is always $1$ for these irreducible curves). Since the dimension of the tangent space depends solely on the stalk $\mathscr{O}_{X,a}\cong \mathscr{O}_{Y,f(a)}$, we know it must be preserved under isomorphisms. We conclude if $f:X_k\to X_l$ is an isomorphism then the origin goes to the origin.
This means $I = I(\{0\})\unlhd A(X_l)$ is sent to $f^*I = I(\{f(0)\})\unlhd A(X_k)$.
Therefore, the blow-up $\tilde{X_k}$ of $X_k$ at the origin, i.e., at $I(\{0\})$, is isomorphic to the blow-up $\tilde{X_l}$ of $X_l$ at the origin, i.e., at $I(\{f(0)\})$.