What a great question! The term "log-log convexity" is indeed used to refer to the property you have described. It pops up in literature on polynomial optimization, and also in a bit of a disguised manner in geometric programming.
Let's look at the latter case, because to be honest that's what I'm personally familiar with. Consider an expression of the form
$$\sum_{k=1}^K c_k x_1^{a_{1k}} x_2^{a_{2k}} \cdot x_n^{a_{nk}}$$
The variables here are $x_i\geq 0$, $i=1,2,\dots, n$, and the fixed parameters are positive $c_k>0$, $k=1,2,\dots K$ and exponents $a_{ij}$, $1\leq i\leq n$, $1\leq j\leq K$. This expression is what is known as a posynomial. In general, it is not convex, nor is it log-convex.
(Side note: if all of the exponents are nonnegative, and $\sum_i a_{ik}\leq 1$ for each $k=1,2,\dots,K$, then this expression is concave. But we want to consider the case where the exponents are not constrained in this way.)
Now suppose I define $x_j = e^{y_j}$, and substitute. This converts the expression to
$$\sum_{k=1}^K c_k e^{a_{1k}y_1} e^{a_{2k}y_2} \cdot e^{a_{nk}y_n}
= \sum_{k=1}^K c_k e^{\sum_{i=1}^n a_{ik}y_i} =
\sum_{k=1}^K \mathop{\textrm{exp}}\left(\textstyle\sum_{i=1}^n a_{ik}y_i+\log c_k\right)$$
Notice how we have depended upon the positivity of the coefficients $c_k$ so we could move then into the exponents; and notice how the exponents are affine functions of $y$.
Now, if I take the logarithm, I get
$$\log \sum_{k=1}^K \mathop{\textrm{exp}}\left(\textstyle\sum_{i=1}^n a_{ik}y_i+\log c_k\right) = f(Ay+\bar{c})$$
where $A\in\mathbb{R}^{n\times k}$ is a collection of the $a_{ij}$ values, $\bar{c}\in\mathbb{R}^K$ is a collection of the $\log c_k$ values, and $f$ is a convex log-sum-exp function
$$f(z) \triangleq \log \sum_{i=1}^K \mathop{\textrm{exp}}(z_k).$$
So in other words, by doing a logarithmic transformation of the variables, and taking the logarithm of the expression, we have arrived at a convex result. What this means is: a posynomial function is a log-log-convex function of its inputs.
(An astute reader will see that the expression was convex before we took that final logarithm. But the log-sum-exp function tends to have better numerical properties the sum-exp function. Also, when $K=1$, as often occurs in practice, the resulting function is linear, and it's useful to exploit that, of course!)
In true geometric programming, posynomials are the only log-log-convex functions considered. Even in generalized geometric programming, the models considered are converted to pure geometric form, so again, posynomials remain the basic nonlinear construct. But it would be entirely possible to consider an entire family of log-log-convex functions here.
One thing I will point out is that standard convexity and log-log-convexity don't mix very well. That is to say, you usually can't mix convex expressions of $x$ and log-convex expressions of $y=\log x$ in the same model, if you wish to preserve the theoretical and practical benefits of convex optimization. (I leave the cases where you can do so to the reader; hint: consider $y\leq \log x$.) That said, you can mix convex and log-convex functions of $y$ in the same model (e.g., sum-exp and log-sum-exp).
One option is to check directly that the definition of a convex function is satisfied.
It's useful to know that any norm on $\mathbb R^n$ is a convex function. Proof: If $x,y \in \mathbb R^n$ and $0 \leq \theta \leq 1$, then
\begin{align*}
\| \theta x + (1 - \theta) y \| & \leq \| \theta x \| + \| (1 - \theta) y \| \\
&= \theta \| x \| + (1 - \theta) \| y \|.
\end{align*}
This shows that the definition of a convex function is satisfied.
When $n = 1$, the $2$-norm is just the absolute value function $f(x) = | x |$. This shows that the absolute value function is convex.
A bunch of other techniques for recognizing convex functions are explained in the book Boyd and Vandenberghe (free online).
Best Answer
Basically Boyd is taking the approach of showing what happens with $\log \det$ in a neighborhood of $t=0$. You're quite correct in that you can make $Z+tV$ into something that's not positive definite, but that isn't really the point here. In fact if you let $t=0.1$, then the sum is still p.d.
What this approach is doing is to show that if you have a line through any $Z$, then you can show that it's concave along that line. That doesn't mean that you have to follow that line out of the domain under scrutiny. It just gives you a way to show convexity locally.
As for $V$ being a "direction": this might be a little confusing until you realize that matrices can be vectors, and that $S^n$ is just a vector space. $Z+tV$ is just the expression of the fact that $Z$ is a vector, and by adding $tV$ (also a vector), you're moving in a line through $Z$. $V$ can be any symmetric matrix, because the important thing is that $Z$ is a member of the domain you're looking at ($S^n_{++}$, the domain of $\log \det$).
So assuming $t$ is small enough, you can answer the question of whether $g$ is concave on a line through $Z$. The fact that $S^n_{++}$ is an open subset of $S^n$ lets us do that.