When can I assume a force to be constant? If I have a force $F(x)$, can I assume it to be constant in any infinitesimal interval such as $Rd\theta$,$ dy \over cos\phi$, $dz$ etc. or can I assume it to only be constant in the interval $[x,x+dx)$? If yes, can you prove it mathematically and if not, which infinitesimal intervals can I assume a force to be constant in & why?
 A: What you want to do is a Taylor Expansion. Assuming that  $F\in C^1(\mathbb{R}^3)$ you have:
$$ F(\mathbf{x})=F(\mathbf{x}_0)+\nabla F(\mathbf{x})|_{\mathbf{x}_0}\cdot(\mathbf{x}-\mathbf{x}_0) + \mathcal{o}(\mathbf{x}-\mathbf{x}_0)$$
So, you can approximate $F$ to be constant only if the first (and higher) order terms are neglectable for your purposes. Here I've showed you a general expansion, but you can write the gradient in any coordinate system (for example spherical) you wish. If it's legitimate to consider only the zeroth order of the expansion usually depends on the explicit form of $F(\mathbf{x})$ and the particular application in which you want to make this approximation.

Let's now consider a one-variable function $F:\mathbb{R}\to\mathbb{R}$, $F\in C^N(\mathbb{R})$. It's then possible to write it using the Taylor theorem:
$$ F(x)=\sum_{k=0}^{N}\frac{d^kF}{dx^k}|_{x_0}\frac{(x-x_0)^k}{k!} +\mathcal{o}((x-x_0)^k)$$
If we put N=1, we get:
$$ F(x)=F(x_0)+\frac{dF}{dx}|_{x_0}\cdot(x-x_0)+\mathcal{o}(x-x_0)$$
That means that you can legitimately consider $F(x)\simeq F(x_0)=\text{constant}$ in the neighborhood of $x_0$ only if the first (and higher) order derivative of $F(x)$ evaluated in $x=x_0$ times $(x-x_0)$ is small enough to be considered neglectable for your purposes.
Here I wrote $F:=F(x)$, but it's exactly the same if $F:=F(\theta)$. Just put $x\mapsto\theta$ (or any other variable you wish) in the above formulas.

Consider now the limit for $\mathbf{x}\to \mathbf{x_0}$, i.e. $\mathbf{x}-\mathbf{x_0}\simeq d\mathbf{x}$. The differential of the function $F(\mathbf{x})$ is its linear increase respect to its variables. It is defined as:
$$dF=\nabla F \cdot d\mathbf{x} =\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz$$
So F is not increasing (i.e. $F$ is constant) when moving along a direction only if the partial derivative in the same direction is zero. You can write this in any coordinate system you want. For example in spherical:
$$dF=\frac{\partial F}{\partial r}dr+\frac{1}{r}\frac{\partial F}{\partial \theta}rd\theta+\frac{1}{r\sin\theta}\frac{\partial F}{\partial \phi}r\sin\theta d\phi$$
So just evaluate the partial derivative and see if it's zero. If that's the case, then you can consider the function constant along that direction.
