Find the ordinary generating function $h(z)$ for a Gambler's Ruin variation. Assume we have a random walk starting at 1 with probability of moving left one space $q$, moving right one space $p$, and staying in the same place $r=1-p-q$. Let $T$ be the number of steps to reach 0. Find $h(z)$, the ordinary generating function.
My idea was to break $T$ into two variables $Y_1,Y_2$ where $Y_1$ describes the number of times you stay in place and $Y_2$ the number of times we move forward or backward one. Then try to find a formula for $P(T=n)=P(Y_1+Y_2=n)=r_n$, but I'm getting really confused since there are multiple probabilities possible for each $T=n$ for $n\geq 3$. Once I have $r_n$ I can then use $h_T(z)=\sum_{n=1}^\infty r_n z^n$, but I'm not sure where to go from here.

I do not have the references to find the generating function of $Y_2$ with me, and it is rather tedious to re-do those computations. I'll skip this part, and only find the generating function of $T$ provided the generating function of $Y_2$ is known. Let $F (z) = E (z^{Y_2})$ be the generating function of $Y_2$.
First, let us write the generating function of $T$ and change the indices of the summation:
$$h (z) = \sum_{N=1}^{+ \infty} P (T = N) z^N = \sum_{N=1}^{+ \infty} \sum_{k=0}^{N-1} P (Y_2 = N-k, Y_1 = k) z^N = \sum_{n=1}^{+ \infty} \sum_{k=0}^{+ \infty} P (Y_2 = n, Y_1 = k) z^{n+k}.$$
Note that, is $Y_2$ is known, the law $Y_1$ is a sum of $Y_2$ i.i.d. random variables of geometric law: after each step, you spend an geometric time of parameter $p+q$ waiting before taking the next step. Since the law of $Y_1$ knowing $Y_2$ is well-known, let us condition over the value of $Y_2$.
$$h (z) = \sum_{n=1}^{+ \infty} P (Y_2 = n) z^n \sum_{k=0}^{+ \infty} P (Y_1 = k | Y_2 = n) z^k.$$
The generating function of a single random variable of geometric law of parameter $p+q$ is:
$$g (z) = \sum_{k=0}^{+ \infty} (1-p-q) (p+q)^k z^k = \frac{1-(p+q)}{1-(p+q)z}.$$
The generating function of the sum of $n$ independent random variables with the same law is $g^n$ (generating functions are really, really nice), so that:
$$\sum_{k=0}^{+ \infty} P (Y_1 = k | Y_2 = n) z^k = g(z)^n = \left( \frac{1-(p+q)}{1-(p+q)z} \right)^n.$$
If we inject this expression into the formula for $h$, we get:
$$h (z) = \sum_{n=1}^{+ \infty} P (Y_2 = n) z^n g(z)^n = F (z g(z)) = F \left( \frac{[1-(p+q)]z}{1-(p+q)z} \right).$$
The last step is to find $F$. But I hope this clears your confusions.
Edit : fixed the confusion between geometric and exponential laws. I also want to stress the point that finding $F$ is feasible, the method is well-known, but this is another matter (besides, I don't think that your problem is there). I prefer to keep this answer rather short, and if necessary I will add something about $F$. Later.
