In general, you shouldn't expect to be able to compute the effective resistance of an electrical circuit using only series and parallel reductions; the class of graphs (called series-parallel graphs) for which this can be done is quite special. There are other transformations for computing resistances such as star-triangle / delta-Y transformations but similar comments hold.
However, resistance computations for metrized graphs are really not so bad. They always boil down to solving a linear system of equations involving the Laplacian matrix of a weighted graph which models the metrized graph. This expository paper of Baker and Faber seems to be a good reference; see Theorem 4 for the relation between the Laplacian operator of a metrized graph and the Laplacian matrix of a weighted graph model.
Since the two points you care about are vertices of the metrized graph, you can proceed as follows. First, write down the Laplacian matrix for a weighted graph model of your metrized graph with the same vertex set $V$ (see Definition 7 of Baker and Faber). This is a symmetric $|V|\times|V|$ matrix $Q$. Let $\delta_j$ be the $|V|$-dimensional vector whose components are all zero except for the $j$th component, which is 1. Then it is possible to find a vector $v$ such that:
(Note that $Q$ is not invertible, so $v$ is only defined up to addition of scalar multiples of the vector $(1,1,1,\dots,1)$).
The effective resistance you want is $r(p,q)=v_p-v_q$; note that this difference is well-defined even if $v$ is not.
You can derive this formally from the properties of $r(p,q)$ (see e.g. section 6 of Baker and Faber), but I wrote it down using the following physical intuition: suppose one unit of current is injected at $p$ ($+\delta_p$) and removed at $q$ ($-\delta_q$), then the effective resistance is equal to the potential difference between $p$ and $q$.
One way to compute $r(p,q)$ without worrying about the kernel of $Q$ is to delete any row and column of $Q$ to get a matrix $\tilde Q$ (now with full rank $|V|-1$) and delete the corresponding rows of $\delta_p,\delta_q$ to get vectors $\tilde\delta_p,\tilde\delta_q$. Then there is a unique solution to
$$\tilde Q\tilde v=\tilde\delta_p-\tilde\delta_q,$$
(throw this into your favorite computer algebra / numerical linear algebra program) and $\tilde v_p-\tilde v_q$ is the effective resistance you want.