Projective $n$-space has a natural open covering by $n+1$ copies $\Bbb A^n_i$ of affine space; we can consider $\Bbb A^n_0\subseteq\Bbb P^n$ to be defined by the map $(a_1^{(0)},\ldots,a_n^{(0)})\mapsto[1:a_1^{(0)}:\ldots:a_n^{(0)}],$ $\Bbb A^n_1\subseteq\Bbb P^n$ defined by $(a_1^{(1)},\ldots,a_n^{(1)})\mapsto[a_1^{(1)}:1:a_2^{(1)},\ldots,a_n^{(1)}],$ and so on.
Given a subvariety $Y\subseteq \Bbb A^n$, to find its projective closure, we first want to choose some identification of $\Bbb A^n$ with an isomorphic subspace of $\Bbb P^n.$ A natural way to do this is via $\Bbb A^n\cong\Bbb A^n_i$ for any choice of $i.$ Then $\overline Y$ will be the Zariski closure of $Y\subseteq\Bbb A^n_i\subseteq\Bbb P^n,$ defined by the homogenization of the defining equations for $Y$ in the coordinates of $\Bbb A^n_i$ with respect to $x^i.$
In your notation, I believe you have chosen $U_0=\Bbb A^n_0.$ But, to answer your first question, it is not necessarily true that $\varphi(Y)=\overline Y;$ instead we identify $Y$ with $\varphi(Y)$ and $\varphi(Y)^{\mathrm{cl}}=\overline Y$, where $\mathrm{cl}$ denotes Zariski closure.
I usually like to think of projective closure as "compactification" of an affine thing, i.e. the affine $Y$ will be Zariski dense in the closed projective $\overline Y.$