Here's the first part. Let $\mu, x$ be some eigenvalue/eigenvector pair of the matrix $Yvv^T$ that you have in the trace. Then by definition:
\begin{align}
\exists x \neq 0, Y vv^T x &= \mu x \\
\lambda v v^T x &= \mu x \text{ (since }v, \lambda \text{ is an eigenvalue/vector of } Y \text{)}\\
\lambda v^T x &= \mu v^T x \text{ (left multiply by } v^T \text{)}\\
(\lambda-\mu)v^Tx & = 0
\end{align}
For this to hold, either $v^Tx=0$, or $v^Tx \neq 0$ and $\lambda=\mu$.
This shows that $Yvv^T$ has eigenvalue $0$ for all $x$ such that $v^Tx=0$, and $\lambda$ for all $x$ such that $v^Tx \neq 0$. The corresponding eigenspaces are therefore $\text{null}(v)$ and $\text{range}(v)$ , which have dimensions $n-1$ and $1$.
Since the trace is just the sum of eigenvalues, repeated by algebraic multiplicity, $$\text{tr}(Yvv^T) = 0 \times (n-1) + \lambda \times 1 = \lambda $$
For the second part (thanks @Hetebrij for pointing this out), we similarly note that the eigenvalue of $I + tvv^T$ is $1$ for $x\in\text{null}(v)$ and $1+t$ for $x\in\text{range}(v)$ (notice $P=vv^T$ is a projection matrix, so $tP$ has eigenvalues $0$ and $t$), thus
\begin{align}
\text{det} (I + tvv^T) &= 1^{n-1} (1+t)^1 = 1+t
\end{align}
As for this choice of $X=I + tvv^T$, I guess it's because it nicely exploits the fact that $Y \nprec 0$ and turns out to be sufficient (we're essentially evaluating $f^*(y)$ along a "line" of $X$s in $\mathbb{S}_{++}^n$, with an end point $I$ and direction vector $vv^T$, so parameter $t$ tells us how far we travel along the line). Anyway the point is to show that $f^*(y)$ is unbounded above for $Y \nprec 0$, and any form of $X \succ 0$ that shows this would work.
