Regarding the weight of the given modular form I am following this course. A question regarding the modular form of weight two that is constructed in lecture 33 emerged. Let me briefly tell you what's going on there.
Let $\mathbb{H}$ denote the upper half-plane. For $\tau \in \mathbb{H}$, define  $L(\tau)=\{n+m\tau: n,m \in \mathbb{Z}\}$. The Weierstrass $\wp-$function associated to $\tau$ is given by 
\begin{equation} \wp_{\tau}(z) = \dfrac{1}{z^2}+\sum\limits_{\omega \in L(\tau) \setminus \{(0,0)\}} \bigg(\dfrac{1}{(z-\omega)^2} - \dfrac{1}{\omega^2}\bigg).
\end{equation}
In what follows, I will omit the subscript $\tau$.
It is well known that this function satisfies the differential equation 
        \begin{equation}
  (\wp'(z))^2=4\wp^3(z)-g_2\wp(z)-g_3,
  \end{equation}
    where $g_2=60 \sum\limits_{\omega \in L(\tau) \setminus \{(0,0)\}} \dfrac{1}{\omega^4}$ and $g_3=140 \sum\limits_{\omega \in L(\tau) \setminus \{(0,0)\}} \dfrac{1}{\omega^6}$.
We may also write the above equation as
\begin{equation}
  (\wp'(z))^2=4(\wp(z)-e_1)(\wp(z)-e_2)(\wp(z)-e_3),
 \end{equation}
where $e_i=e_i(\tau)$. Note that the values of $z$ at which $\wp'$ vanishes correspond precisely to the values of $z$ for which $\wp(z)=e_i$ for some $i\in\{1,2,3\}$. It can be shown that the only (simple) zeros of $\wp'$ in the parallelogram with vertices $0$, $1$, $\tau$, $1+\tau$ are $\frac{1}{2}$, $\frac{\tau}{2}$, $\frac{1+\tau}{2}$. So we may put  $e_1=\wp(\frac{1}{2})$, $e_2=\wp(\frac{\tau}{2})$, and $e_3=\wp(\frac{1+\tau}{2})$. It can also be shown that all these quantities viz. $e_1,$ $e_2,$ $e_3$ are distinct for any $\tau \in \mathbb{H}$ and that they are holomorphic functions of $\tau$ on the upper half-plane.
Now define the function $\lambda: \mathbb{H}\rightarrow \mathbb{C}$ by $\lambda(\tau)=\dfrac{e_3(\tau)-e_2(\tau)}{e_1(\tau)-e_2(\tau)}$, which is holomorphic on its domain and never takes the values $0$ and $1$. It can be shown that this function is invariant under the subgroup of $PSL(2,\mathbb{Z})$ congruent $\mod2$, i.e., under the kernel of the homomorphism $PSL(2,\mathbb{Z}) \rightarrow PSL(2,\mathbb{Z}/2\mathbb{Z})$ defined in the obvious way.
The professor calls thus constructed function $\lambda$ a modular form of weight two (which also follows from the title of the lecture), but he never explains what the weight of a modular form is. (In one of the preceding lectures, he mentions, as far as I remember, that modular forms or modular functions are automorphic functions that are invariant under the unimodular group $PSL(2,\mathbb{Z})$ or a subgroup of the unimodular group.) So the question is why $\lambda$ is a modular form of weight two? It is not obvious for me if I accept the definition of a modular form of weight $k$ from Wikipedia. 
 A: A modular form is an holomorphic function $f\colon\mathbb{H}\to\mathbb{C}$ satisfying some growth conditions and the following modular property:
$$f(\frac{az+b}{cz+d})=(cz+d)^kf(z)$$ for 
$$ \begin{pmatrix}
a & b \\
c & d
\end{pmatrix} 
\in \Gamma\subset SL(2,\mathbb{R}),
$$
and $\Gamma$ is a discrete group. The integer $k$ is called the weight of the modular form (but different authors, as Serre in his book "A course in arithmetic", uses to write $2k$ instead of $k$ to denote the weight).
In this case, your weight is $k=2$ and the group $\Gamma$ you are considering is the group of matrices in $PSL(2,\mathbb{Z})$ with $a,d$ odd and $b,c$ even.
To verify that for such matrices the above relation holds for $k=2$, you can try with a direct computation, which is not difficult. 
First you have to write explicitly, for $\tau\in\mathbb{H}$
$$\lambda(\tau)=\frac{\mathcal{P}(1/2)-\mathcal{P}(\tau/2)}{\mathcal{P}((\tau+1)/2)-\mathcal{P}(\tau/2)} = \frac{\sum_{\omega\in L(\tau)}{\biggl(\frac{1}{(1/2-\omega)^2}-\frac{1}{(\tau/2-\omega)^2}\biggr)}}{\sum_{\omega\in L(\tau)}{\biggl(\frac{1}{((\tau +1)/2-\omega)^2}-\frac{1}{(\tau/2-\omega)^2}\biggr)}} = 
\sum_{\omega\in L(\tau)}{\frac{((\tau+1)/2-\omega)^2\bigl((\tau/2-\omega)^2-(1/2-\omega)^2\bigr)}{(1/2-\omega)^2\bigl((\tau/2-\omega)^2-((\tau +1)/2-\omega)^2\bigr)}} =
\sum_{\omega\in L(\tau)}{\frac{((\tau+1)/2-\omega)^2((\tau+1)/2-2\omega)((\tau -1)/2)}{(1/2-\omega)^2((2\tau+1)/2-2\omega)(-1/2)}}.
$$ 
Now take a matrix in $PSL(2,\mathbb{Z}/2\mathbb{Z})$ and obtain the new element $$\frac{a\tau+b}{c\tau+d}\in\mathbb{H}.$$
Now you have to do exactly the same computations as above (note that the fractions involved have always non zero denominator due to the fact that $a,d$ are odd and $b,c$ even) and you will find that an expression which is equivalent to the previous one (after you do the sum over the lattice) times the factor $(c\tau+d)^2.$ This clearly proves the statement.
