I'm currently reading these notes and they are quite good, many exercises inside for becoming familiar with blow-ups.
Interesting example of blow-up often comes as a linear system. Consider a family of curves of degree $d$ in the plane : $\lambda F + \mu G = 0$ with $F,G$ polynomials of degree $d$. To each point of $\mathbb P^2$ we can associate a unique $(\lambda : \mu) \in \mathbb P^1$ such that $p \in Z(\lambda F + \mu G)$ except when $p \in Z(F) \cap Z(G)$ (this locus is called the base-point of the linear system). For obtaining a well-defined map, you need to blow up the plane at $d^2$ points : you obtain then a well-defined map $Bl_Y(\mathbb P^2) \to \mathbb P^1$ and you can study your linear system which is a fibration (possibly with singular fibers).
The book " 4-manifolds and Kirby calculus" has also a nice section about blow-up.
Finally, any notes online about algebraic geometry also have a chapter about blow-up and classical examples.
(This is rather a comment that an answer, but I don't have enough space in comments).
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 $(x)$.
Best Answer
This is how I like to think about proper transform. This is somewhat loosely speaking be warned.
Say you are blowing up $Z\subset X$, to get a space $\pi:\hat X\to X$. For a point, say $p$, in the base space, if $p\notin Z$ there is a unique preimage $\hat p$ in $\hat X$ which 'lives above' $p$. On the other hand, if $p\in Z$, then living above $p$ a big chunk of the exceptional divisor, namely the fiber over $p$.
Now suppose you are blowing up $p\in \mathbb C^2$, then the preimage of $p$ is an entire $\mathbb P^1$. On the other hand, say we blowup a line $\mathbb P^1 \cong Z\subset\mathbb P^3$, the entire exceptional divisor is $E\cong\mathbb P^1\times \mathbb P^1$, and the fiber over any point $p$ sees one fiber in $E$.
Great, now in the latter situation, what if you had a curve $C$ passing through $Z$, say intersecting it at a single point. If you just take the set theoretic inverse image you get a curve with a $\mathbb P^1$ attached to it in $\hat X$. This is called the total transform and isnt really what we want a lot of the time.
For this reason, you can follow the following mantra to get the 'right' curve $\hat C$ living above $C$ in $\hat X$. Take $\pi^{-1}(C)$, and 'throw out the exceptional stuff', i.e. set theoretically remove the intersection with the exceptional divisor. Now take closure. This is the proper transform in this case, and you get more what you expect. Instead of this exceptional fiber attached to $\hat C$.
Working out a few examples, like the one above can be quite helpful. You can do this in coordinates even! Take your favourite line inside $\mathbb P^3$ and blow it up, check what the inverse image is, try to figure out what the proper transform looks like.