= Solution
A <modular form> of integer weight $k$ for a finite-index subgroup $\Gamma\leq SL_2(\mathbb Z)$ is a <holomorphic function> on the <complex upper half-plane> satisfying $f|_k\gamma=f$ for $\gamma\in\Gamma$ and <holomorphic> at every <cusp of a modular group>. Explicitly, for each $\sigma\in SL_2(\mathbb Z)$ choose a positive integer $h_\sigma$ with $\sigma T^{h_\sigma}\sigma^{-1}\in\Gamma$. Then $f|_k\sigma$ is periodic with period $h_\sigma$, and its expansion in $q_{h_\sigma}=e^{2\pi iz/h_\sigma}$ must have no negative powers. Equivalently it must be bounded as $\operatorname{Im}z\to\infty$, uniformly in one period. This definition applies to noncongruence subgroups too. If $-I\in\Gamma$, nonzero forms require even weight.
Put $d=[SL_2(\mathbb Z):\Gamma]$ and let $f\ne0$. Form the product over left cosets
$$
F(z)=\prod_{\gamma\in\Gamma\backslash SL_2(\mathbb Z)}(f|_k\gamma)(z).
$$
Changing a representative does not affect a factor. Right multiplication permutes the cosets, so the product is a nonzero level-one <modular form> of weight $kd$. It is <holomorphic> on the upper half-plane and at infinity, since every factor is bounded there by cusp holomorphy.
Let $h$ be the least positive integer with $T^h\in\Gamma$, and write $f=\sum_{n\geq0}b_nq_h^n$. If $r$ is its first nonzero index, the $h$ distinct cosets $\Gamma T^j$, $0\leq j<h$, give the factors $f(z+j)$. Their product has leading term a nonzero constant times $e^{2\pi irz}=q^r$. All remaining factors are bounded at infinity. Hence $F(z)=O(e^{-2\pi r\operatorname{Im}z})$, so its level-one order satisfies $v_\infty(F)\geq r$. Notice that using just the single factor $f$ would only give $r/h$; the entire translation orbit is needed.
The permitted <valence formula for the modular group> bounds $v_\infty(F)\leq kd/12$, because all its other weighted zero orders are nonnegative. Thus $r\leq kd/12$. With $R=\lfloor kd/12\rfloor$, the map $f\mapsto(b_0,\ldots,b_R)$ is injective: a nonzero form in its kernel would have $r\geq R+1$. This proves the <dimension bound for modular forms on a finite-index subgroup>:
$$
\boxed{\dim M_k(\Gamma)\leq\lfloor kd/12\rfloor+1\leq1+kd/12
\quad(k\geq0).}
$$
For the dilation assertion, the original PDF states $N_1D\mid N$, a product-divisibility condition, rather than the TeX's chain $N_1\mid D\mid N$. Use the PDF condition. Put $A=\operatorname{diag}(D,1)$. If $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma_0(N)$, then
$$
A\gamma A^{-1}=\begin{pmatrix}a&Db\\c/D&d\end{pmatrix}\in\Gamma_0(N_1),
$$
since $N_1D\mid c$. Therefore, for $g(z)=f(Dz)$, either direct substitution or the <slash operator for modular forms> gives $g|_k\gamma=g$. It is <holomorphic> on the upper half-plane.
To verify every cusp, take $\sigma\in SL_2(\mathbb Z)$ and choose $\tau\in SL_2(\mathbb Z)$ taking infinity to the rational cusp $A\sigma\infty$. Then $B=\tau^{-1}A\sigma$ is rational upper triangular, with positive ratio of its two diagonal entries. Consequently
$$
(f|_k A)|_k\sigma=(f|_k\tau)|_k B
$$
is a nonzero constant factor times $f|_k\tau$ evaluated at an affine map $z\mapsto uz+v$ with $u>0$. The imaginary part tends to infinity, where $f|_k\tau$ is bounded. Thus $g|_k\sigma$ is bounded too. Together with its periodicity this proves cusp holomorphy by the <Riemann removable singularity theorem>. Hence the <oldform by argument dilation> satisfies
$$
\boxed{f\in M_k(\Gamma_0(N_1)),\ N_1D\mid N
\ \Longrightarrow\ f(Dz)\in M_k(\Gamma_0(N)).}
$$
Finally $\Gamma_0(3)$ has index four: reduction modulo three gives a transitive $SL_2(\mathbb F_3)$ action on the four points of $\mathbb P^1(\mathbb F_3)$, and $\Gamma_0(3)$ is the inverse image of the stabilizer of infinity. Reduction is surjective, since elementary determinant-one matrices generate $SL_2(\mathbb F_3)$ and lift integrally. The dimension bound gives $\dim M_4(\Gamma_0(3))\leq2$.
The normalized <Eisenstein series> $E_4(z)$ and the oldform $E_4(3z)$ both belong to this space. Their <Fourier expansion of a normalized Eisenstein series> starts as
$$
E_4(z)=1+240q+O(q^2),\qquad E_4(3z)=1+240q^3+O(q^6).
$$
A linear relation between them has zero constant term only if its two scalars sum to zero; its $q$ coefficient then forces both scalars to vanish. They are independent, proving the <weight-four modular forms on Gamma 0 3> description
$$
\boxed{\dim M_4(\Gamma_0(3))=2,\qquad
\{E_4(z),E_4(3z)\}\text{ is a basis}.}
$$
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