Direct telescopic proof for sum of $1^2+2^2+...+n^2$
Noting that $$ (n+1)^2-n^2=2n+1, (n+1)^3-n^3=3n^2+3n+1 $$ one has $$ n=\frac12\bigg[(n+1)^2-n^2\bigg]-\frac12,n^2=\frac13\bigg[(n+1)^3-n^3\bigg]-n-\frac13. $$ So $$ n^2=\frac13\bigg[(n+1)^3-n^3\bigg]-n-\frac13=\frac13\bigg[(n+1)^3-n^3\bigg]-\frac12\bigg[(n+1)^2-n^2\bigg]+\frac16 $$ and hence \begin{eqnarray} \sum_{k=1}^nk^2&=&\sum_{k=1}^n\left[\frac13\bigg[(k+1)^3-k^3\bigg]-\frac12\bigg[(k+1)^2-k^2\bigg]+\frac16\right]\\ &=&\frac13\bigg[(n+1)^3-1\bigg]-\frac12\bigg[(n+1)^2-1\bigg]+\frac n6\\ &=&\frac16n(n+1)(2n+1) \end{eqnarray}
Use binomial coefficients. Write $k^2=2\binom{k}{2}+\binom{k}{1}$ so $\sum_{k=1}^nk^2=\sum_{k=1}^n(a_k-a_{k-1})$ with $$a_n=2\binom{n+1}{3}+\binom{n+1}{2}=\frac13 n(n^2-1)+\frac12 n(n+1)=\frac16 n(n+1)(2n+1).$$
I quite endorse the other answers. Proffering the following alternative approach to producing correct "telescoping" functions for I find it kinda neat.
Your telescoping argument used the fact that for $f(x)=x(x+1)/2$ we have the identity $$ f(x)-f(x-1)=x.\tag{1} $$ The goal is to find another function $F(x)$ satisfying a similar identity $F(x)-F(x-1)=x^2$ instead of $(1)$.
If we integrate both sides of the identity $(1)$ we get the identity $$ \begin{aligned} \frac12x^2&=\int_{t=0}^xt\,dt\\ &=\int_{t=0}^x(f(t)-f(t-1))\,dt\\ &=\int_{t=0}^xf(t)\,dt-\int_{t=-1}^{x-1}f(t)\,dt\\ &=\int_{t=0}^xf(t)\,dt-\int_{t=0}^{x-1}f(t)\,dt-\int_{t=-1}^0f(t)\,dt\\ &=g(x)-g(x-1)+\frac1{12}, \end{aligned} \tag{2} $$ where $$ g(x)=\int_{t=0}^xf(t)\,dt=\frac16x^3+\frac14x^2. $$ Two problems here. We got $\dfrac12x^2$ instead of $x^2$ as the difference. But it is easy to fix that by multiplying everything with two. The other problem is that "integration constant $C=1/12$".
To deal with that pesky $1/12$ let me introduce a correction term $\epsilon(x)=x/12$ with the property $$ \epsilon(x)-\epsilon(x-1)=\frac1{12}.\tag{3} $$ Let's then define the function $$ F(x)=2(g(x)+\epsilon(x))=\frac13x^3+\frac12x^2+\frac16x. $$ Taking into account equations $(2)$ and $(3)$ we get the desired identity $$ F(x)-F(x-1)=x^2.\tag{4} $$ This telescopes the same way your telescoping with $f(x)$ did. Observe that $F(0)=0$. Of course, we have the expected factorization $$ F(x)=\frac16x(2x+1)(x+1). $$
It is hopefully clear how to get a formula for the sum of cubes by calculating appropriate integrals of $F(x)$, and produce a function $G(x)$ with the properties $G(0)=0$ and $$ G(x)-G(x-1)=x^3. $$ Rinse and repeat to your heart's content.