Asymptotics for the sums from the inclusion-exclusion principle

Many unsolved asymptotics problems can be written as inclusion-exclusion sums. There is no general method for solving them.

However, in your case the terms decrease in absolute value and nearly form a geometric series. This is an easy case.

If $t_k$ is the $k$-term, then $t_{w+i}\approx (-1)^i (j/n)^i t_w$, so $$c \sim \frac{t_w}{1+j/n}.$$ To get rigorous bounds, you can use the fact that the partial sums of an alternating series with absolutely decreasing terms alternate above and below the full sum.

Another way to approach this from the generating function perspective. Notice that $$\sum_{k=l}^u (-1)^k [x^k]\, f(x) = [x^u]\,\frac{f(-x)}{1-x} - [x^{l-1}]\,\frac{f(-x)}{1-x},$$ and so the question reduces to studying the asymptotic of the coefficients of $\frac{f(-x)}{1-x}$.

In your example, we have $$(j-k+1)\binom{n-k+1}{j-k+1} = (n-j+1)\,[x^{j-k}]\,(1-x)^{-(n+2-j)}$$ and thus \begin{split} c &=\sum_{k=w}^{j} (-1)^{k-w} (j-k+1)\binom{n-k+1}{j-k+1} \\ &= (n-j+1)(-1)^{j-w} \sum_{t=0}^{j-w} (-1)^t\,[x^t]\,(1-x)^{-(n+2-j)} \\ &= (n-j+1)(-1)^{j-w}\,[x^{j-w}]\,\frac{1}{(1+x)^{n+2-j}(1-x)} \\ &= (n-j+1)\,[x^{j-w}]\,\frac{1}{(1-x)^{n+2-j}(1+x)}. \end{split}

Now, noticing the partial fraction decomposition $$\frac{1}{(1-x)^m(1+x)} = \frac{2^{-m}}{1+x} + \sum_{i=0}^{m-1} \frac{2^{-(i+1)}}{(1-x)^{m-i}},$$ we obtain the formula: $$c=(n-j+1)\left(\frac{(-1)^{j-w}}{2^{n+2-j}} + \sum_{i=0}^{n+1-j} \frac{1}{2^{i+1}}\binom{n+1-i-w}{j-w}\right).$$ Here all terms in the sum are positive and thus are more amenable to asymptotic analysis. For example, we can deduce $$\frac{n-j+1}{2}\binom{n+1-w}{j-w} \lesssim c \lesssim (n-j+1)\binom{n+1-w}{j-w}.$$