Kirchhoff's current law is justified by conservation of charge.
But conservation of charge means that in a closed system, the amount of charge doesn't change, not that it can't accumulate.
So, what do they mean by saying that KCL is a consequence of conservation of charge?
But conservation of charge means that in a closed system, the amount of charge doesn't change, not that it can't accumulate.
If charge is conserved, and negative charge accumulates somewhere in the system, then positive charge must accumulate somewhere else in the system.
Now you have negative charge in one place and positive charge somewhere else, so there will be an electric field between the two. That is, you have formed a capacitor.
But if you wanted to include a capacitor in your circuit you would have drawn a capacitor in your schematic (I'm talking about the circuit that you use for analyis, not necessarily the one you use to construct a physical circuit). Since you didn't draw a capacitor in your schematic, you are stating that there is no significant charge accumulation.
And yes, real wires have parasitic capacitance to other nearby wires and to ground --- that's just another way of saying that charge does accumulate in the circuit to some degree. You can do some analysis or modeling and find out that the values of the parasitic capacitance (aka the amount of accumulated charge) is very often small enough to be negligible. Or you can use the analysis/modeling results to include the parasitic capacitance in your circuit model.
Another way to say this is, in the symbology of schematic diagrams, we define the wire symbol (lines connecting components) to represent equipotential regions with no accumulated charge. And we define the capacitor symbol as the way in which we represent the possibility of charge accumulation.
Without any assumption on the regime (stationary, transient, periodic,...), KCL comes from electric charge continuity applied to circuit nodes, i.e. (approximately) zero-dimensional volumes.
In general,
$$\dfrac{d}{dt} \int_{V} \rho + \oint_{\partial V} \mathbf{j} \cdot \hat{\mathbf{n}} = 0 \ ,$$
as you shrink the volume to a "small volume", the volume decreases as $\sim L^3$, while its boundary decreases as $\sim L^2$. Taking the limit for $L \rightarrow 0$, the leading term is the boundary term, and the leading term of the equation reads,
$$0 = \oint_{\partial V} \mathbf{j} \cdot \hat{\mathbf{n}} = \sum_k I_k \ ,$$
i.e. KCL.
Of course, if you take a finite-dimenisonal volume $V$, without any other assumptions, you can store net electric charge inside the volume, and the integral form of the charge balance gives you
$$\dfrac{d}{dt}{Q}_V = \sum_k I_k \ .$$
You're right: there's more to it than charge conservation.
Consider a water tank with pipes connected to it below the water level. Despite conservation of mass, the sum of mass flow rates for water entering does not necessarily equal that for water leaving.
Why is it different for electric charge at a junction? Not, I'd say, because there's no electrical analogue for the tank, but because of repulsion between like charges: that's what stops accumulation (or deficit) of electrons at a junction.
To offer a more mathematical explanation, this follows from the continuity equation
$$\frac{\partial \rho}{\partial t} + \nabla\cdot \mathbf{J}=0. $$
A wire is macroscopically neutral, and therefore, macroscopically, $\rho=0$. Alternatively, you can consider $\partial_t\rho=0$ by assuming no charge accumulates anywhere in the wire. Therefore, the continuity equation is just $\nabla \cdot \mathbf{J}=0$. Now you invoke Gauss' theorem
$$\int_{V}\nabla \cdot \mathbf{J} \ dV = \oint_{S}\mathbf{J}\cdot d\vec{S} \to \oint_{S}\mathbf{J}\cdot d\vec{S} =0.$$
But the surface integral of $\mathbf{J}$ is the current $I$. In this case, the surface is a closed surface, what that means is that the net current through a closed surface is zero, $I_\mathrm{net}=0$. This is KCL: the sum of currents through a node, the net current, is zero.
