The cosine similarity between two vectors $a$ and $b$ is just the angle between them
$$\cos\theta = \frac{a\cdot b}{\lVert{a}\rVert \, \lVert{b}\rVert}$$
In many applications that use cosine similarity, the vectors are non-negative (e.g. a term frequency vector for a document), and in this case the cosine similarity will also be non-negative.
For a vector $x$ the "$z$-score" vector would typically be defined as
$$z=\frac{x-\bar{x}}{s_x}$$
where $\bar{x}=\frac{1}{n}\sum_ix_i$ and $s_x^2=\overline{(x-\bar{x})^2}$ are the mean and variance of $x$. So $z$ has mean 0 and standard deviation 1, i.e. $z_x$ is the standardized version of $x$.
For two vectors $x$ and $y$, their correlation coefficient would be
$$\rho_{x,y}=\overline{(z_xz_y)}$$
Now if the vector $a$ has zero mean, then its variance will be $s_a^2=\frac{1}{n}\lVert{a}\rVert^2$, so its unit vector and z-score will be related by
$$\hat{a}=\frac{a}{\lVert{a}\rVert}=\frac{z_a}{\sqrt n}$$
So if the vectors $a$ and $b$ are centered (i.e. have zero means), then their cosine similarity will be the same as their correlation coefficient.
TL;DR Cosine similarity is a dot product of unit vectors. Pearson correlation is cosine similarity between centered vectors. The "Z-score transform" of a vector is the centered vector scaled to a norm of $\sqrt n$.
