Your second equation seems a bit off. In the curve of equation:
$$ y^2 + cy = x^3 + ax + b $$
in a binary field $\mathbb{F}_{2^m}$, to add point $P_1 = (x_1,y_1)$ to point $P_2 = (x_2,y_2)$, resulting in point $P_3 = (x_3,y_3)$, then the two equations are:
\begin{eqnarray*}
x_3 &=& \lambda^2 + x_1 + x_2\\
y_3 &=& \lambda (x_1+x_3) + y_1 + c
\end{eqnarray*}
where
$$ \lambda = \frac{y_1+y_2}{x_1+x_2} $$
In projective coordinates, the point $(x,y)$ is represented by the triplet $(X,Y,Z)$ such that $x = X/Z$ and $y = Y/Z$. It is important to realize that the projective coordinates are not unique; $(X,Y,Z)$ and $(\mu X, \mu Y, \mu Z)$ represent the same point, for any non-zero $\mu$. If we have projective coordinates for $P_1$ and $P_2$, then we get:
\begin{eqnarray*}
\lambda &=& \frac{\frac{Y_1}{Z_1}+\frac{Y_2}{Z_2}}{\frac{X_1}{Z_1}+\frac{X_2}{Z_2}} \\
&=& \frac{\frac{Y_1 Z_2 + Y_2 Z_1}{Z_1 Z_2}}{\frac{X_1 Z_2 + X_2 Z_1}{Z_1 Z_2}} \\
&=& \frac{Y_1 Z_2 + Y_2 Z_1}{X_1 Z_2 + X_2 Z_1}
\end{eqnarray*}
Then:
\begin{eqnarray*}
\frac{X_3}{Z_3} &=& \frac{(Y_1 Z_2 + Y_2 Z_1)^2}{(X_1 Z_2 + X_2 Z_1)^2} + \frac{X_1}{Z_1} + \frac{X_2}{Z_2} \\
\frac{Y_3}{Z_3} &=& \frac{Y_1 Z_2 + Y_2 Z_1}{X_1 Z_2 + X_2 Z_1} \cdot \left(\frac{X_1}{Z_1} + \frac{(Y_1 Z_2 + Y_2 Z_1)^2}{(X_1 Z_2 + X_2 Z_1)^2} + \frac{X_1}{Z_1} + \frac{X_2}{Z_2}\right) + \frac{Y_1}{Z_1} + c
\end{eqnarray*}
The goal is to compute without making divisions; thus, we need to define $Z_3$ such that all these inconvenient fractions disappear. One possible solution, given the equations above, is to set:
$$ Z_3 = Z_1 Z_2 (X_1 Z_2 + X_2 Z_1)^3 $$
from which we deduce:
\begin{eqnarray*}
X_3 &=& (X_1 Z_2 + X_2 Z_1)(Z_1 Z_2 (Y_1 Z_2 + Y_2 Z_1)^2 + (X_1 Z_2 + X_2 Z_1)^2(X_1 Z_2 + X_2 Z_1)) \\
&=& Z_1 Z_2 (X_1 Z_2 + X_2 Z_1) (Y_1 Z_2 + Y_2 Z_1)^2 + (X_1 Z_2 + X_2 Z_1)^4
\end{eqnarray*}
and some even more complex -- but without any division -- expression for $Y_3$ (that I am too lazy to compute).
So you get the two main points:
- yes, to get projective coordinates, you "just" replace $x$ with $X/Z$ and so on;
- but ultimately you still get some choice in the expression for $Z_3$, since projective coordinates are not unique. You thus choose an expression which minimizes overall computational cost, which usually means "the simplest expression which avoids divisions".
Since you have a choice in $Z_3$, you do not necessarily end up with the same formulas as other people, and it is not necessarily mean that your formulas are wrong.
