There's an exact sequence $$0\to \pi^*\Omega^1_X\to \Omega^1_{\widetilde{X}}\to i_*\Omega^1_{E/Y}\to 0$$ where $i:E\to \widetilde{X}$ is the inclusion of the exceptional divisor. This corresponds to the fact that pulling back differential forms gives you differential forms which are constant along the fibers. Checking that this is exact may be done from looking at stalks: stalks of pullbacks are easy to compute, and stalks of direct images along closed immersions are easy to compute, plus each $\Omega^1$ is locally free on it's appropriate space, as everything involved here is smooth.
Taking $\mathcal{Hom}(-,\mathcal{O}_{\widetilde{X}})$ of this exact sequence, we get the long exact sequence $$ 0\to \mathcal{Hom}(i_*\Omega^1_{E/Y},\mathcal{O}_{\widetilde{X}}) \to \mathcal{Hom}(\Omega^1_\widetilde{X},\mathcal{O}_{\widetilde{X}}) \to \mathcal{Hom}(\pi^*\Omega^1_X,\mathcal{O}_{\widetilde{X}}) \to \mathcal{Ext}^1(i_*\Omega^1_{E/Y},\mathcal{O}_{\widetilde{X}})\to \cdots$$
and as $\mathcal{Hom}(-,\mathcal{O}_X)$ is exact on locally free sheaves, we see that $ \mathcal{Ext}^i(\pi^*\Omega^1_X,\mathcal{O}_{\widetilde{X}})$ and $\mathcal{Ext}^i(\Omega^1_\widetilde{X},\mathcal{O}_{\widetilde{X}})$ both vanish for $i>0$. As $i_*\Omega^1_{E/Y}$ is torsion, it's sheaf hom in to $\mathcal{O}_\widetilde{X}$ vanishes, and so all that's left to do is to identify $\mathcal{Ext}^1(i_*\Omega^1_{E/Y},\mathcal{O}_\widetilde{X})$. Replacing $i_*\Omega^1_{E/Y}$ with a locally free resolution locally isomorphic to $0\to\mathcal{O}_\widetilde{X}^d(-E)\to\mathcal{O}_\widetilde{X}^d\to 0$, we see the identification $\mathcal{Ext}^1(i_*\Omega^1_{E/Y},\mathcal{O}_{\widetilde{X}})\cong i_*T_{E/Y}(E)$. So our long exact sequence reduces to $$0\to T_{\widetilde{X}}\to \pi^*T_X\to i_*T_{E/Y}(E)\to 0.$$
This is the correct relationship between $T_X$ and $T_\widetilde{X}$. It's completely independent of $Y$ as long as $Y$ is a smooth subvariety.
The correct setting for an equation like you ask for in (2) to hold is with the canonical sheaf of a blowup. Per Griffiths and Harris, for example, $\omega_{\widetilde{X}} \cong \pi^*\omega_X\otimes \mathcal{O}((c-1)E)$ where $c$ is the codimension of $Y$ in $X$ and $\omega_X=\bigwedge^{\dim{X}} \Omega^1_X$. Translating this in to tangent sheaves by dualizing and using the fact that dualizing commutes with taking determinant bundles (plus that $(\mathcal{O}(mE))^*=\mathcal{O}(-mE)$), we see that we should have $\bigwedge^nT_\widetilde{X}\cong (\bigwedge^nT_X)((1-c)E)$. There is no relation between $c$ and $\deg Y$.
There are two things to keep in mind. First, we always need to fix an ample line bundle to speak about stability. Secondly, Intersection Theory is made a way that you can work (at least if $X$ is smooth) on the Grothendieck group $K_0(X)$, so one might consider locally free resolution.
Fix an ample line bundle $H$. To define the degree of any coherent sheaf $F$, you can consider a locally free resolution
$$0\to F_m \to \cdots \to F_1 \to F_0 \to F$$
and define $\det(F)=\prod \det(F_i)^{(-1)^i}$. Then you define
$$\deg(F) = c_1(\det(F))\cdot H^{\dim(X)-1}$$
This definition agrees with the one you state in the case of maximal dimension (see http://www.math.harvard.edu/~yifei/tensor_char_zero.pdf Lemma $1.20$).
Example: If you consider an effective divisor $D\in Pic(X)$, you have the exact sequence $0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X \to \mathcal{O}_D \to 0$, thus $c_1(\mathcal{O}_D)=D$ (as a cycle) and thus $\deg_H(\mathcal{O}_D)=D\cdot H^{\dim X -1} =\deg_H(D)$ (counting multiplicities and taking the sum on irreducible components).
_**Example in a blow-up:*__ Now consider the simple example of $\widetilde{X}=Bl_{p}\mathbb{P}^2 \to \mathbb{P}^2$, the blowup of $\mathbb{P}^2$ at a point $p$. If I take a the strict transform $\widetilde{C}$ of an effective divisor $C\in Pic(\mathbb{P}^2)$, it is still an effective divisor, and $\widetilde{C}=\pi^*C-rE$ where $r$ is the multiplicity of $C$ at $p$. Then there is a resolution
$$0 \to O_{\widetilde{X}}(-D) \to O_{\widetilde{X}} \to O_D \to 0$$
and we obtain once again $c_1(\mathcal{O}_{\widetilde{C}})=\widetilde{C}$. Thus if you fix your ample divisor $T=m\pi^*H - E$ on $\widetilde{X}$, you get
\begin{eqnarray}
\deg_T(\mathcal{O}_{\widetilde{C}}) &=& \widetilde{C}\cdot T\\
&=& (\pi^*C-rE)\cdot(m\pi^*H-E)\\
&=&m\deg_H(C)-r.
\end{eqnarray}
Example for vector bundle :
Consider $\pi : \widetilde{X} \to X$ the blow-up of a smooth projective scheme along a smooth projective subscheme and write $E$ the exeptional divisor. Fix an ample divisor $H$ on $\widetilde{X}$. For a vector bundle $V$ of rank $r$ on $X$, choose a resolution
$$0 \to V_n \to \cdots \to V_0 \to V \to 0,$$
thus $\deg_H(V)= c_1\left(\prod (\det V_i)^{(-1)^{i}}\right)\cdot H^{\dim X-1} = \sum (-1)^i c_1(\det V_i)\cdot H^{\dim X-1}$.
Now define $\widetilde{V}=\pi^* V \otimes O_{\widetilde{X}}(-lE)$ for some $l\in\mathbb{Z}$. As pullback is exact when apply to locally free sheaves, you have the resolution
$$0 \to \pi^* V_n \to \cdots \to \pi^* V_0 \to \pi^* V \to 0.$$
Fix the ample line bundle $T=m\pi^* H - E$ on $\widetilde{X}$. As $c_1(\widetilde{V})=c_1(\pi^*(V)\otimes \mathcal{O}_{\widetilde{X}}(-lE) = c_1(\pi^*V)-rlE$, we obtain
\begin{eqnarray}
\deg_T (\widetilde{V}) &=& (c_1(\pi^* V)-rlE)\cdot T^{\dim X -1}\\
&=&\sum (-1)^ic_1( \det \pi^* V_i)(m\pi^* H^{\dim X-1})+rlE^2 \\
&=& m\deg_H(V)+rlE^2
\end{eqnarray}
In more general settings, I don't know if things go so well. You need to find a locally free resolution of the strict transform, which might be harder as pullback is not exact in general.
Best Answer
1) The blowup of a sheaf of ideals $\mathcal{I}\subset \mathcal{O}_X$ is the relative Proj of the blowup algebra $\bigoplus_{m\geq 0} \mathcal{I}^m$. This is a definition, and we're just applying it to the affine case here where we replace $\mathcal{O}_X$ by $A$ and $\mathcal{I}$ by $I$. Choosing a set of generators $f_1,\cdots,f_r$ for $I$ gives you a surjection from $A^r\to I$, which turns in to a surjection of graded rings $\operatorname{Sym}(A^r)\to \bigoplus_{m\geq 0} I^m$, which corresponds to a closed immersion of their Projs by the general properties of that construction.
2) Every algebraic geometry book which covers blowing up should have a proof. For instance, Griffiths and Harris pg 184, Stacks 02OS, or this MSE question for a low-dimensional example (which generalizes).
3) This is just dualizing the statement of 2). Saying "$E$ corresponds to $\mathcal{O}(-1)$" means that $\mathcal{O}(E)\cong\mathcal{O}(-1)$, and dualizing gives $\mathcal{O}(-E)\cong\mathcal{O}(1)$ as requested.
4) Serre vanishing is not actually used in the proof: it's only meant to point out that the claim $\mathcal{O}_{\widetilde{X}}(-nE)\cong I_Y^n$ holds for any $X,Y$ once $n\gg 0$. For the specific situation of the proof, we are able to use the argument involving the exact sequence from the (now doubly) linked post to show this for $n\geq 1$ directly.
5) You've missed a $(n)$ here: the correct statement should be that global sections of $\pi_*\mathcal{O}_{\Bbb P(A^r)}(n)$ are exactly $\operatorname{Sym}^n(A^r)$.
Both of these claims follow from the property that for nice rings $R$, the sheaf $\mathcal{O}(n)$ on $\operatorname{Proj} R$ has global sections exactly $R_n$, the degree-$n$ portion of $R$. See for instance the discussion in Stacks 01QG.
In the first case, the ring $R$ is $\operatorname{Sym}(A^r)$, with graded components $\operatorname{Sym}^n(A^r)$, and in the second, the ring is $\bigoplus_{m\geq 0} I^m$ with graded component $I^n$, and we use the isomorphism $\mathcal{O}_{\widetilde{X}}(n)\cong\mathcal{O}_{\widetilde{X}}(-nE)$.