The statement, as claimed, is false.

Let $p = 2, N = 11$, and let $f_0$ be the unique normalised eigenform in $S_2(\Gamma_0(11))$; and set $f(\tau) = f_0(8\tau)$. Then $f \in M_2(\Gamma_0(Np^3))$, but $U_p(f) = f_0(4\tau)$ is not in $M_2(\Gamma_0(Np))$.

However, $U_p(f)$ is in $M_2(\Gamma_0(Np^2))$. The correct general statement is that for $r \ge 2$, $U_p$ maps $M_2(\Gamma_0(Np^r))$ to $M_2(\Gamma_0(Np^{r-1}))$ (and likewise for forms of any weight $k$, not just weight $2$). But in general it doesn't go all the way from level $\Gamma_0(Np^r)$ to $\Gamma_0(Np)$.

PS. You asked for an outline of the proof. The idea is to check that (under the hypotheses of my last paragraph) the double coset $\Gamma_0(Np^r) \begin{pmatrix} 1 & 0 \\ 0 & p \end{pmatrix} \Gamma_0(Np^r)$ is actually invariant under right-translation by $\Gamma_0(Np^{r-1})$, not merely by $\Gamma_0(Np^r)$. (This kind of argument is very important in Hida theory.)

The statement, as claimed, is false.

Let $p = 2, N = 11$, and let $f_0$ be the unique normalised eigenform in $S_2(\Gamma_0(11))$; and set $f(\tau) = f_0(8\tau)$. Then $f \in M_2(\Gamma_0(Np^3))$, but $U_p(f) = f_0(4\tau)$ is not in $M_2(\Gamma_0(Np))$.

However, $U_p(f)$ is in $M_2(\Gamma_0(Np^2))$. The correct general statement is that for $r \ge 2$, $U_p$ maps $M_2(\Gamma_0(Np^r))$ to $M_2(\Gamma_0(Np^{r-1}))$ (and likewise for forms of any weight $k$, not just weight $2$). But in general it doesn't go all the way from level $\Gamma_0(Np^r)$ to $\Gamma_0(Np)$.

PS. You asked for an outline of the proof. The idea is to check that (under the hypotheses of my last paragraph) the double coset $\Gamma_0(Np^r) \begin{pmatrix} 1 & 0 \\ 0 & p \end{pmatrix} \Gamma_0(Np^r)$ is actually invariant under right-translation by $\Gamma_0(Np^{r-1})$, not merely by $\Gamma_0(Np^r)$. (This kind of argument is very important in Hida theory.)

EDITED TO ADD. One can check that the composite of $U_p$ with the twisted map $M_2(\Gamma_0(Np^{r-1})) \to M_2(\Gamma_0(Np^r))$, $f(\tau) \mapsto f(p\tau)$, is multiplication by a power of $p$. So $U_p$ is surjective as a map $M_2(\Gamma_0(Np^r)) \to M_2(\Gamma_0(Np^{r-1}))$, answering the slightly more refined question in the title of Adithya's post.

The statement, as claimed, is false.

Let $p = 2, N = 11$, and let $f_0$ be the unique normalised eigenform in $S_2(\Gamma_0(11))$; and set $f(\tau) = f_0(8\tau)$. Then $f \in M_2(\Gamma_0(Np^3))$, but $U_p(f) = f_0(4\tau)$ is not in $M_2(\Gamma_0(Np))$.

However, $U_p(f)$ is in $M_2(\Gamma_0(Np^2))$. The correct general statement is that for $r \ge 2$, $U_p$ maps $M_2(\Gamma_0(Np^r))$ to $M_2(\Gamma_0(Np^{r-1}))$ (and likewise for forms of any weight $k$, not just weight $2$). But in general it doesn't go all the way from level $\Gamma_0(Np^r)$ to $\Gamma_0(Np)$.

The statement, as claimed, is false.

Let $p = 2, N = 11$, and let $f_0$ be the unique normalised eigenform in $S_2(\Gamma_0(11))$; and set $f(\tau) = f_0(8\tau)$. Then $f \in M_2(\Gamma_0(Np^3))$, but $U_p(f) = f_0(4\tau)$ is not in $M_2(\Gamma_0(Np))$.

However, $U_p(f)$ is in $M_2(\Gamma_0(Np^2))$. The correct general statement is that for $r \ge 2$, $U_p$ maps $M_2(\Gamma_0(Np^r))$ to $M_2(\Gamma_0(Np^{r-1}))$ (and likewise for forms of any weight $k$, not just weight $2$). But in general it doesn't go all the way from level $\Gamma_0(Np^r)$ to $\Gamma_0(Np)$.

PS. You asked for an outline of the proof. The idea is to check that (under the hypotheses of my last paragraph) the double coset $\Gamma_0(Np^r) \begin{pmatrix} 1 & 0 \\ 0 & p \end{pmatrix} \Gamma_0(Np^r)$ is actually invariant under right-translation by $\Gamma_0(Np^{r-1})$, not merely by $\Gamma_0(Np^r)$. (This kind of argument is very important in Hida theory.)