How can I derive the forward differencing formula using Taylor series in order to approximate derivative of a function at some point $x$?
I don't see how we can link the two formulas. I came across a solution where:
$$ f(x) = f(a) + f^{\prime}(a)(x-a) + \frac{f^{\prime\prime}(a)(x-a)^{2}}{2!} $$
$f(a)$ was replaced by $f(x_{i})$ and $f(x)$ was replaced by $f(x_{i+1})$ but I don't see why?
Next, $f^{\prime}(x_{i})$ was taken out and it was equal to $$f^{\prime}(x_{i}) = \frac{f(x_{i+1})}{h} + \frac{f(x_{i})}{h} + \frac{f^{\prime\prime}(x_{i})h}{2}$$
where
$$ h = x_{i+1} -x{i} $$ and the last term is the error term.
Does this suffice as an answer?
No, it is not sufficient. You need to combine \begin{align} f(x_i)&=f(x_i)\\ f(x_{i+1})&=f(x_i)+f'(x_i)h+\frac12f''(x_i)h^2+\frac16f'''(x_i)h^3+...\\ f(x_{i+2})&=f(x_i)+2f'(x_i)h+2f''(x_i)h^2+\frac43f'''(x_i)h^3+...\\ \end{align} to eliminate $f(x_i)$ and the second derivative terms on the right side.