I am trying to solve the following exercise; in more general situations also I do not know how to address the problem of finding the equations for the image of a Riemann surface under the embedding induced by a very ample divisor.
We consider a complex torus $E=\mathbb{C}/L$ and we denote $p_0=[0] \in E$. I have to describe a basis of $L(4p_0)$ and find equations for the image of $X$ in $\mathbb{P}^3$ under the embedding induced by the divisor $4p_0$. I have more or less no idea where to start.
We have done something similar in class in the case of $3p_0$: we have observed that there must be some non constant function in $L(2p_0)$ and that this function, say $f$ must have a pole of order $2$ at $p_0$ and no other poles and must be even; up to rescaling and subtracting a constant we can WLOG suppose that a local representative of $f$ is given by $f(z)=1/z^2+\alpha z^2+\mathrm{h.o.t.}$ for some $\alpha \in \mathbb{C}$, hence a basis for $L(2p_0)$ is given by $\{1, f\}$. Taking (a multiple of) the derivative of $f$ we get a meromorphic function $g=-\frac{1}{2}f'$ which is odd and has a pole of order $3$ at $p_0$; $\{1, f, g\}$ is a basis of $L(3p_0)$ and using Laurent expansion we get that $g^2-f^3\in L(2p_0)$, hence $g^2-f^3=af+b$ for some constants. So the image of the embedding $\phi(z)=[1:f(z):g(z)]$ is in the zero locus of the homogenized polynomial $h(x_1, x_2, x_3)=x_1x_3^2-x_2^3+ax_1^2x_2+bx_1^3$ and we can actually prove that $\phi(X)=Z(h)$.
Now in the case of the exercise the most natural thing seemed considering the function $h= -\frac{1}{3} g'$ and form a basis of $L(4p_0)$ as $\{1, f, g, h\}$, but as for the equations for $\phi(X)$ I have no ideas.
Take $\{1, f, g, f^2\}$ as a basis for $L(4 p_0)$ and define \begin{align*} \varphi: \mathbb{C}/L &\to \mathbb{P}^3\\ z &\mapsto [T_0 : T_1 : T_2 : T_3] = [1 : f(z) : g(z) : f(z)^2] \, . \end{align*} Since $f^2$ has a pole of order $4$ at $0$, then dividing through the above expression by $f(z)^2$ we find $$ 0 \mapsto [1/f(0)^2 : 1/f(0) : g(0)/f(0)^2 : 1] = [0:0:0:1] \, . $$ Note that the image of $\phi$ lies on $T_1^2 = T_0 T_3$, and from the equation $g^2 = f^3 + af + b$, it also lies on $$ T_2^2 = T_1 T_3 + a T_0 T_1 + b T_0^2 \, . $$ (Here we have used $f^3 = f \cdot f^2 = T_1 T_3$. I think you may have run into trouble by using $f^3 = T_1^3$ instead.)
Let \begin{align*} h_1 &= T_1^2 - T_0 T_3\\ h_2 &= T_2^2 - (T_1 T_3 + a T_0 T_1 + b T_0^2) \end{align*} and let $C: h_1 = h_2 = 0$. We claim that $C$ is the image of $\varphi$. Certainly $\operatorname{img}(\varphi) \subseteq C$, so it suffices to show the reverse inclusion. We begin by showing that $C$ is a (irreducible) curve. On the open subset $U_0$ where $T_0 \neq 0$, let $u = T_1/T_0, v = T_2/T_0, w = T_3/T_0$. Dehomogenizing $h_1$ and $h_2$, set $C$ is described by \begin{align*} u^2 &= w\\ v^2 &= uw + a u + b \end{align*} on $U_0$. Eliminating $w$, then $C \cap U_0$ is isomorphic to $C': v^2 = u^3 + a u + b$, which is an irreducible curve. When $T_0 = 0$, then $T_1^2 = 0$ from $h_1$ so $T_1=0$, and then $T_2^2 = 0$ from $h_2$, so $T_2 = 0$. Thus $[0:0:0:1]$ is the only point of $C_0$ on the hyperplane $T_0 = 0$. Therefore $C$ is an irreducible curve.
Thus $\varphi: E \to C$ is a map between irreducible projective curves, hence is either constant or surjective. Since $\varphi$ is not constant, then it is surjective and thus $\operatorname{img}(\varphi) = C$.