$$\frac1{x(x+1)}+\frac1{(x+1)(x+2)}+\frac{p-2}{x(x+2)}=\frac{p}{x(x+2)}.$$

Also, $$\frac1{(x+1)(x+2)}-\frac1{(x+3)(x+4)}={\frac {4x+10}{ \left( x+2 \right) \left( x+1 \right) \left( x+3
\right) \left( x+4 \right) }}.$$ So, over $\mathbb{Z}_5,$ the only $x$ for which this is defined is $x=0$ and there the numerator is $0.$ However the two are independent in $\mathbb{Z}_5(x).$

I think that if the $a_i$ and $b_i$ are all unequal then the expressions are linearly independent. In fact

If $\{\frac1{(x-a_i)(x-b_i)}\}_{i=1}^k$ is dependent in the over some field, but all subsets are independent, then in the list $a_1,b_1,a_2,b_2,\cdots , a_k,b_k$ every entry appears at least twice.

Consider $$\sum_1^k\frac{c_i}{(x-a_i)(x-b_i)}$$ where the $a_i,b_i,c_i$ are $3k$ parameters. This is a rational function with numerator a polynomial of degree $2k-2.$ The coefficient of $x^{k-2}$ is $\sum c_i$ so one could arrange for the leading coefficient, and others, to be zero. The question, as I will answer it, is if all the coefficients could be zero, i.e. if the numerator could be identically zero as a formal polynomial. Alternately, one might wonder if the numerator evaluates to zero for all $x$ in the field (not of the form $x=a_i$ or $x=b_i.$) The example above shows an instance of this. However I am not considering that, and for $k$ small with respect to the order of the field this can't happen.

I will assume that the parameters are chosen so that $a_1$ is not repeated among the other $a_i$ and $b_i$ and that $c_1 \neq 0.$ Without loss of generality we can take $a_1=0$ and $c_1=1.$ Then the numerator has constant term $\prod_{i=2}^k(a_ib_i)\neq 0.$

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