= Solution
Choose holomorphic coordinates centered at $p$. Locally, the <blowup of a complex manifold at a point> is
$$
\widetilde{\mathbb C^n}=\{(z,[\ell])\in\mathbb C^n\times\mathbb{CP}^{n-1}:z\in\ell\},
$$
with $\sigma(z,[\ell])=z$; away from $p$ this is an isomorphism, so it glues to $X\setminus\{p\}$. The <exceptional divisor> is $E=\sigma^{-1}(p)\cong\mathbb{CP}^{n-1}$.
The <proper transform> is the closure of $\sigma^{-1}(Y\setminus\{p\})$. If $p\notin Y$, it is isomorphic to $Y$. If $p\in Y$, the holomorphic <implicit function theorem> supplies coordinates in which $Y=\{z_1=0\}$. In the blowup chart with $z_i=t$ and $z_j=tu_j$ for $j\ne i$, every chart with $i\ne1$ describes the proper transform by $u_1=0$, while the chart $i=1$ does not meet it. These are smooth coordinate hypersurfaces, so $\widetilde Y$ is smooth.
For a divisor $D$, the <line bundle associated to a divisor> $\mathcal O(D)$ consists locally of <meromorphic function>[meromorphic functions] $f$ such that $(f)+D\geq0$. Pulling back a local defining function for $Y$ shows that its divisor is
$$
\sigma^*Y=\widetilde Y+mE,
$$
where $m$ is the <order of vanishing> at $p$ of a local defining function for $Y$. Therefore
$$
\sigma^*\mathcal O(Y)\cong\mathcal O(\widetilde Y+mE).
$$
Because $Y$ is smooth, $m=1$ when $p\in Y$ and $m=0$ when $p\notin Y$.
Applying the definition with $D=E$ gives directly
$$
H^0(\widetilde X,\mathcal O(E))\cong\{f\text{ meromorphic on }\widetilde X:(f)+E\geq0\}.
$$
The map sends a section to its local meromorphic coefficient relative to the canonical meromorphic section of $\mathcal O(E)$; the divisor inequality is exactly the condition that these coefficients define a holomorphic section, and the inverse construction is local multiplication by that canonical section.
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