Find a Taylor series for $f(z)=z\cos(2z)$ around $z_0= \pi$
I found that the Taylor series for $f(z)=\cos(2z)$ is $\sum_{n=0}^{\infty}\frac{(-1)^n4^n}{(2n)!}(z-\pi)^{2n}$. So do I just multiply in $z$ for $f(z)=z\cos(2z)$ to get $\sum_{n=0}^{\infty}\frac{(-1)^n4^n}{(2n)!}(z-\pi)^{2n}z$?
That is a power series for the function. You can do arithmetic with power series just like with the functions. Be careful if you need to evaluate such expressions though, as doing this can have surprising effects on the radius of convergence (not in your case since this function is analytic).
However, it's not the Taylor polynomial. You can tell this because Taylors polynomials hae the form $$\sum \frac{f^{(n)}(c)}{n!}(x-c)^n$$ which your formula doesn't have. An easy way to tell is to note that you have both $z$ and $z-\pi$ in your expression. You can modify it to obtain the desired for though, by multiplying by $\pi+(z-\pi)$ (instead of multiplying by $z$) and leaving the second term grouped.