I'm not an expert here, but I'll try. Let $X$ be a proper irreducible variety and $K$ the field of meromorphic function on $X$. Let $v : K^* \to A$ be a valuation, where $A$ is a totally ordered abelian group. Recall that the valuation ring $R$ is $v^{-1}(A_{\geq 0})$, with maximal ideal $\mathfrak{m} = v^{-1}(A_{> 0})$. So $R$ is a local ring with fraction field $K$. By the valuative criterion of properness, we get a map $\mathrm{Spec} \ R \to X$; the image of the closed point is called the "center" of $v$. We denote the center by $Z$.

I'll start with the simplest cases, and move to the more general. The easiest case is that $X$ is defined over an algebraically closed field $k$, the point $Z$ is a closed point of $X$, and $A$ is $\mathbb{Z}$. Then the completion of $R$ at $\mathfrak{m}$ is isomorphic to $k[[t]]$; choose such an isomorphism. If $x_1$, $x_2$, ... $x_N$ are coordinates near $z$ then this isomorphism identifies $z_i$ with a power series $\sum x_i^j t^j$.

Intuitively, we can think of $(x_1(t), x_2(t), \ldots, x_N(t))$ as giving an analytic map from a small disc $D$ into $X$, taking the origin to $Z$. (This interpretation makes rigorous sense if we are working over $\mathbb{C}$, and the power series $x_i$ converge somewhere.) The image of this map is the "branch" which Zariski speaks of. The valuation can be thought of as restricting a function to this branch and seeing to what order it vanishes at $Z$. Notice that, if there is a polynomial relation $f(x_1(t), \ldots, x_N(t))=0$ then we should have $v(f)=\infty$ and $v(1/f)= - \infty$. If you don't allow this then you need to require that the $x_i$ be algebraically independent; intuitively, this is the same as saying that the branch does not lie in any polynomial hypersurface. Since Zariski allows "algebraic" branches, I assume he IS permitting this and has some appropriate convention to deal with it.

Now, what if $A$ is a subgroup of $\mathbb{R}$, but is no longer discrete? Then $R$ is going to embed in some sort of Puiseux field. The details here can be subtle, but the intuition should be that the $x_i(t)$ can have real exponents. For example, if $K=k(x,y)$, and $v(f(x,y))$ is the order
of vanishing of $f(t, t^{\sqrt{2}})$ at $t=0$, then we can think of $v$ as the order of vanishing along the branch $(t, t^{\sqrt{2}})$. Maybe that is what Zariski means by a transcendental branch??

Suppose now that $Z$ is the generic point of an $s$-dimensional variety. Let $L$ be the field of functions on $Z$ and suppose, for simplicity, that $L$ and $K$ have the same characteristic. If $A=\mathbb{Z}$ then $R$ embeds in $L[[t]]$. Again, we can take coordinates $x_1$, ..., $x_N$ and write them as power series. How to think of these power series? One way is to think of them as giving a map $U \times D$ into $X$, where $D$ is again a small disc and $U$ is a dense open subset of $X$. The image is the $(s+1)$-dimensional branch which Zariski discusses. The valuation is to restrict to this branch and work out the order of vanishing of this restriction along $U \times \{ 0 \}$.

Now, all of this is discussing the case where $A$ embeds (as an ordered group) in $\mathbb{R}$. In general, of course, there are more valuations. For example, let $A=\mathbb{Z} \times \mathbb{Z}$ ordered lexicographically. Let $v:k[x,y] \to A$ send a polynomial to its lowest degree monomial, and extend this to a valuation on $k(x,y)$. The way I would think of that is that we have a little disc near $(0,0)$, and a little curve passing through $(0,0)$ along the $x$-axis. So our valuation is to, first, restrict to the curve and the order of vanishing at the point and, second, to restrict to the surface and take the order of vanishing along the curve. So here we have a flag of branches, not just one. I'm not sure why your Zariski quote doesn't discuss this possibility.