Consider the following simple situation: We work over the ring $R=\mathbb{Z}_p[[t]]$, over which we consider a rank $2$ free module $M$ with basis $(e,f)$. On $M$, we define a flat (topologically nilpotent) connection with log poles by the formula $$\nabla: f\mapsto e\otimes\frac{dt}{t}.$$ In other words, we have $t\frac{df}{dt}=e$.

Now, fix some finite extension $K/\mathbb{Q}_p$ with ring of integers $\mathcal{O}_K$ and choose two non-units $q,q'\in\mathcal{O}_K$. Identifying these elements with continuous maps $q,q':R\to\mathcal{O}_K$, we get two specializations $M_q$ and $M_{q'}$ of $M$ over $K$. We can now try to perform ($p$-adic) parallel transport of sections from $M_{q'}$ to $M_q$ using the usual Taylor series: $$f(q')\mapsto\sum_{i=0}^\infty\frac{d^if}{dt^i}(q)\frac{(q'-q)^i}{i!}.$$

Here, by $f(q)$, I of course mean the pull-back of sections $q^*f$.

It is easily checked that the above series simplifies to $$f(q)+\left(-\sum_{i=1}^\infty\frac{(1-q'q^{-1})^i}{i}\right)e(q).$$ The series in parentheses is immediately seen to be the one for $\log(q'q^{-1})$. This, as is well known, will converge only if $q'q^{-1}$ is a principal unit (that is, it is integral and has residue class $1$).

But, of course, we can go ahead and *fix* some branch of the $p$-adic logarithm, say, the one such that $\log(p)=0$. And then we can simply interpolate this isomorphism, which is given above by parallel transport in the cases where $q'q^{-1}$ is a principal unit, to *all* pairs $(q,q')$ by the formula $$f(q')\mapsto f(q)+\log(q'q^{-1})e(q).$$

**Question**: What is in fact going on here? Is there an intrinsic description of this interpolation (without the choice of bases, etc.)? In particular, can there be some general result along the following lines: for any $R$-module $M$ with flat connection with log poles, and any choice of $p$-adic logarithm, there is an interpolation as above of parallel transport between specializations at $K$-valued points, with certain properties? I'm sure somewhere in Dwork, Katz, et al, there is something about this, but it seems to have evaded me so far.