For every $n>0$, let $U_n=1+p^n\mathbb{Z}_p$ be the group of units in $\mathbb{Z}_p$ which are $\equiv1\pmod{p^n}$. I know the proof of following theorem:
Let $x\in U_n$ with $n>0$ if $p\neq2$ and $n>1$ if $p=2$. There is a unique morphism of groups $f_x:\mathbb{Z}_p\to U_n$ such that if $a=(a_r)_{r>0}$ (with $a_r\in\mathbb{Z}/p^r\mathbb{Z}$) and $x=(x_r)_{r>0}$ (with $x_r\in U_n/U_{n+r}$), then $f_x(a)=(x_r^{a_r})_{r>0}$. If $x\notin U_{n+1}$, then $f_x$ is an isomorphism.
From this how can I prove the following corollary:
The group $U_1$ is isomorphic to $\mathbb{Z}_p$ for $p\neq2$ and the group $U_2$ is isomorphic to $\mathbb{Z}_2$ for $p=2$.
Also, I am not able to comprehend the point of showing isomorphism between a group and a ring.
Edit: Can following fact (derived from main theorem) be helpful:
$U_n$ is a free $\mathbb{Z}_p$-module of rank 1.
The key sentence:
Suppose that $p$ is odd. Then $x = 1 + p\in U_1$, so by the proposition you stated, there is a homomorphism of groups $$ f_x : \Bbb Z_p\to U_1. $$ Moreover, because $1 + p\not\in U_2$, $f_x$ is an isomorphism (again by the proposition), so $\Bbb Z_p\cong U_1$ (as groups). If $p = 2$, the proof is exactly the same, taking $x = 1 + 2^2\in U_2\setminus U_3$.
To address the edit: $U_n$ is a free $\Bbb Z_p$-module of rank one, this means exactly that $U_n\cong\Bbb Z_p$ as groups (plus some extra data about the $\Bbb Z_p$-action on the group).