How does one transform a sum into a telescoping sum?

You are essentially asking for a general method to sum any finite or infinite series (or indefinite sum). Correct? There is no such method. For certain special families of series, there do exist methods. For instance, summing of polynomial sequences, that is, $\sum_n n^k$ using Bernoulli numbers. An important special case similar to telescoping series is using Wilf–Zeilberger pairs which requires a computer algorithm.

In general, there is no suitable technique for all cases. Some series don't exactly have telescoping components that are easy to track. One example of types of series that may have a telescoping component are hypergeometric series, whre we wish to sum anything of the form

$$\sum_{k=0}^n\frac{(a_1)_k...(a_p)_k}{(b_1)_k...(b_q)_k}$$

where the notation $(a)_k$ denotes the ascending factorial function. Hypergeometric functions are any functions that can be written in such a fashion, along with the term $x^k/k!$ and the upper limit is taken as infinite. Finding the telescoping series for these requires the use of Gosper's Algorithm, or its modified version Zielberger's Algorithm. Here is an example of how to use Gosper's algorithm.

Suppose that we want a closed form for the sum

$$S_m=\sum_{k=0}^m(-1)^k\binom{n}{k}$$

where $m\leq n$. To evaluate

$$S_m=\sum_{k=0}^ma_k$$

we notice that $S_m-S_{m-1}=a_m$. This, the sum $S_m$ is a sort of antiderivative of the discrete sum. As such, $S_m$ is sometimes called the indefinite sum of $a_k$. Without going into detail, we wish to write the ratio of the terms such that

$$\frac{a_{k+1}}{a_k}=\frac{p_{k+1}}{p_k}\frac{q_k}{r_k}$$

for some polynomials $p_k,q_k$, and $r_k$. One condition that we must enforce (for complicated reasons) is that if $(k+\alpha)$ divides $q_k$ and $(k+\beta)$ divides $r_k$, then $\alpha-\beta-1$ cannot be a positive integer. This is always possible by multiplying appropriate terms to the polynomial $p_k$ until this is true. For our problem, we have

$$\frac{a_{k+1}}{a_k}=\frac{(-1)^{k+1}\binom{n}{k+1}}{(-1)^k\binom{n}{k}}=\frac{k-n}{k+1}$$

Here, $p_k=1$, $q_k=k-n$, and $r_k=k+1$, which satisfy our hypotheses. Gosper's Algorithm states that if a hypergeometric telescoping term exists, then it must take the form

$$S_{k-1}=\frac{r_{k-1}a_ks_k}{p_k}$$

for some polynomial $s_k$ (again, the reasoning is complicated). The polynomial $s_k$ is the solution to

$$p_k=q_ks_{k+1}-r_{k-1}s_k$$

Throwing in our equations, we have

$$1=(k-n)s_{k+1}-ks_k$$

If we assign $s_k=-1/n$, which is a polynomial in $k$ of degree $0$, then this satisfies the polynomial condition. Thus, we have

$$S_{k-1}=(-1)^{k+1}\frac{k\binom{n}{k}}{n}=(-1)^{k+1}\binom{n-1}{k-1},\;\;\;\;S_k-S_{k-1}=a_k$$

An important rule when dealing with hypergeometric terms is if a factorial ever becomes negative, we simply regard it as $0$. Our final answer is

$$\sum_{k=0}^m(-1)^k\binom{n}{k}=S_m-S_{-1}=(-1)^{m}\binom{n-1}{m}$$

Plenty of other hypergeometric sums and series can be established, such as $k!$, $1/(k^2-1)$, etc. More generally, however, we can't always use this, but we can find recursion relations between similar series. For example, Let's consider the sum

$$S=\sum_{k=0}^n\binom{n}{k}^2$$

Gosper's Algorithm will yield no polynomial $s_k$ that we need. Instead, we seek to use Gosper's Algorithm on the modified sum

$$S=b_0\sum_{k=0}^n\binom{n}{k}^2+b_1\sum_{k=0}^{n+1}\binom{n+1}{k}^2$$

for some coefficients $b_n$ to be specified later. Applying the term ratio gives

$$\frac{a_{k+1}}{a_k}=\frac{b_0\binom{n}{k+1}^2+b_1\binom{n+1}{k+1}^2}{b_0\binom{n}{k}^2+b_1\binom{n+1}{k}^2}=\frac{b_0(k-n)^2+b_1(n+1)^2}{b_0(k-n-1)^2+b_1(n+1)^2}\frac{(k-n-1)^2}{(k+1)^2}$$

We assign $p_k=b_0(k-n-1)^2+b_1(n+1)^2$, $q_k=(k-n-1)^2$, and $r_k=(k+1)^2$. The polynomial $s_k$ satisfies

$$b_0(k-n-1)^2+b_1(n+1)^2=(k-n-1)^2s_{k+1}-k^2s_k$$

Without going into detail, the quantity $s_k=2k-3(n+1)$ satisfies the equation with $b_0=-2(2n+1)$ and $b_1=n+1$. We now have

$$S_{k-1}=\frac{k^2(-2(2n+1)\binom{n}{k}^2+(n+1)\binom{n+1}{k}^2)(2k-3(n+1))}{-2(2n+1)(k-n-1)^2+(n+1)^3}$$

We then have partial sums of

$$-2(2n+1)\sum_{k=0}^m\binom{n}{k}^2+(n+1)\sum_{k=0}^m\binom{n+1}{k}^2=\frac{(m+1)^2(-2(2n+1)\binom{n}{m+1}^2+(n+1)\binom{n+1}{m+1}^2)(2(m+1)-3(n+1))}{-2(2n+1)(m-n)^2+(n+1)^3}$$

Letting $m$ be large enough such that the factorial terms vanish, we have that the indefinite sum vanishes, and

$$\sum_{k=0}^{n+1}\binom{n+1}{k}^2=\frac{2(2n+1)}{n+1}\sum_{k=0}^n\binom{n}{k}^2$$

This yields a recursion relation for the sum in question. For $n=0$, the sum is $1$. Using the recursion, we get the next sums are $2,6,20,$ etc. These are precisely the central binomial coefficiants, so we finally have

$$\sum_{k=0}^n\binom{n}{k}^2=\binom{2n}{n}$$.

This algorithm may be employed many times over on many different sums, where the sums may also be infinite.

You are looking for $$ a(n) = r(n) - r(n - 1) $$ or, what is more standardly adopted $$ a(n) = s(n + 1) - s(n) = \Delta s(n) $$

Then $s(n)$ is called the Antidelta or Indefinite Sum of $a(n)$, and same as for integrals some functions have a closed form for the Antidelta, and some don't.

You can find more details in the indicated link.