If, given that the frictional force is constant, one wants to show that the work done against friction is proportional to the length of the path, would this line of reasoning be correct?
We can use the work integral $\int_C \bf{F}$ $\cdot d\bf{x}$ over the path $C := \bf{g}$$(t)$, $t \in [a, b]$. The integral will then be equal to $\int_a^b \bf{F(g}$$(t)) \cdot \bf{g}$$'(t)dt$. We can then define $\|F\|=K$ for some constant $K>0$ and write the integral as $\int_a^b K$$\|\bf{g}$$'(t)\|dt = Kl$, where $l$ denotes the length of the curve $C$.
However, I'm somewhat hesitant, because it is only shown that if $\bf{F}$ is a constant force, the work done by it is proportional to the length of the path.
How does one show that the force is indeed necessarily frictional in this integral? We know that $\bf{F}$ works against the velocity vector $\bf{x}$$'(t)$. Does this mean that the integral should be written as $\int_a^b \bf{F(-g}$$'(t)) \cdot -\bf{g}$$''(t)dt$?
It makes no sense to ask whether the force is "frictional in this integral". The adjective "frictional" describes the physical origin of the force, which is irrelevant to this calculation. One cannot "show that the force is frictional", at least not at this level; to make physical sense of that statement, one would have to model the entire system and see whether the force is caused by dissipative or conservative processes. As long as you're merely talking about the value of $\mathbf F$ and not its origin, there's nothing more to say once you know $\mathbf F$ has constant magnitude and points in the direction opposite the velocity. The formulation that "$\mathbf F$ works against the velocity vector $\mathbf x'(t)$" sounds somewhat unusual to my ears. The last integral doesn't have dimensions of work, it's work per time, and since $\mathbf g''$, the acceleration, is proportional to $\mathbf F$ via $\mathbf F=m\mathbf a$, you're basically integrating the square of the force there, which has no physical significance. The calculation in the first part of the question is basically correct, but I'd mention the changing direction of $\mathbf F$; it's hard to tell from the calculation whether you understand this aspect or merely forgot to include the cosine of the angle.