The first question has nothing to do with geometry, and all to do with linear algebra; in particular the interplay between $V,V^*, V^{**}$ when $V$ is a finite-dimensional vector space (over any field $\Bbb{F}$, doesn't even have to be $\Bbb{R}$). So, that's the situation we shall focus on.
I hope you know that if $\dim V <\infty$, then $\dim V = \dim V^* = \dim V^{**}$, so the spaces are all isomorphic. What's really nice is that $V$ and $V^{**}$ are canonically isomorphic: the map $\iota:V \to V^{**}$ defined by setting for all $v\in V, \omega \in V^*$, $[\iota(v)](\omega) := \omega(v)$ is easily seen to be linear and injective (for example using a basis); then by rank-nullity theorem it follows $\iota$ is actually a linear isomorphism.
Suppose $v\in V$, and $\rho:V^* \to \Bbb{F}$ is such that for all $\omega \in V^*$,
\begin{align}
\rho(\omega) &= \omega(v) \in \Bbb{F}
\end{align}
Then, if you unwind the definition of $\iota$, we see that $\rho(\omega) = \omega(v) = [\iota(v)](\omega)$. Since this is true for all $\omega$, we have that $\rho = \iota(v)$, and this is an equality of elements in $V^{**}$ (i.e it's an equality of $(1,0)$-tensors on $V$). Equivalently, we can rewrite this as $v =\iota^{-1}(\rho)$, and this is now a proper equality of elements in $V$ (and in Lecture $3$, I believe he spends some time trying to explain that $V\cong V^{**}$ when trying to explain why every vector "is" or rather "can be regarded" as a $(1,0)$ tensor).
What usually happens is that in the finite-dimensional case, since the isomorphism $V\cong V^{**}$ using $\iota$ is natural, we simply treat the spaces as being equal, $V=V^{**}$. Of course, set-theoretically, these are different spaces, but whenever we have such natural isomorphisms, it (sometimes) becomes rather cumbersome to keep having to distinguish the spaces. It's kind of like trying to distinguish the spaces $\Bbb{R}\times \Bbb{R}\times \Bbb{R}$ vs $\Bbb{R}^2\times \Bbb{R}$ vs $\Bbb{R}\times \Bbb{R}^2$ vs $\Bbb{R}^3$. Set theoretically, these are different objects, but in most circumstances we just call all of these $\Bbb{R}^3$, and rather than saying "there is a bijective correspondence between the four spaces", we simply say "the four spaces are equal".
In your case, the $\rho$ is just the filled in curvature tensor $R(\cdot, Z,X,Y)$ (if you wish, evaluate everything at a point $p\in M$, then the vector space is $V=T_pM$), while the $v$ is $\nabla_X\nabla_YZ-\nabla_X\nabla_YZ-\nabla_{[X,Y]}Z$. Once again, if you want to be super precise about things, then for each $p\in M$, let $\iota_p: T_pM \to (T_pM)^{**}$ be the canonical isomorphism; then
\begin{align}
\iota_p\bigg((\nabla_X\nabla_YZ)(p)-(\nabla_X\nabla_YZ)(p)-(\nabla_{[X,Y]}Z)(p)\bigg) &= R_p(\cdot, Z(p), X(p), Y(p)).
\end{align}
But, like I said, especially in this finite-dimensional situation, there's no point (once you understand the isomorphism) in trying to keep track of it (because with some practice it should be easy enough to figure out where exactly it goes).
Another way of describing the isomorphism $\iota$ is as follows. Given any vector space $V$, we can always define the "evaluation map" $\text{ev}:V \times V^* \to \Bbb{F}$ by setting $\text{ev}(v,\omega):= \omega(v)$. Why is it called the evaluation map? Because it's purpose is to literally evaluate the given element of $V^*$ on the given element of $V$ to produce a field element. This is easily verified to be a bilinear map.
Sometimes, this is often called the "duality pairing" and is denoted using angle brackets $\langle \cdot, \cdot \rangle$, but it should not be confused with an inner product, because an inner product usually requires real or complex scalar field and is a map $V\times V \to \Bbb{R}$ or into $\Bbb{C}$.
As a result of being bilinear, it induces two linear maps. The first is the mapping $V\to V^{**}$ given by $v\mapsto \text{ev}(v,\cdot)$, and the second is the mapping $V^* \to V^*$ given by $\omega \mapsto \text{ev}(\cdot, \omega)$. The first mapping is precisely the map $\iota$ which I described above, while the second mapping is simply the identity on $V^*$ so it's not interesting.
Just to drive home the point of what $\iota$ does, note that we can always evaluate covectors $\omega$ on a vector $v$ to get a field element $\omega(v)\in \Bbb{F}$. What $\iota$ allows you to do is associate to $v$, an element $\iota(v)$, which can eat covectors to produce a field element $\iota(v)[\omega]:= \omega(v) \in \Bbb{F}$. Now, since $\iota:V\to V^{**}$ is an isomorphism, what this allows us to do is to be slightly sloppy with notation and not write $\iota$ at all in our formulas, and say that "a covector can act on a vector to yield a scalar", and also that "a vector can act on a covector to yield a scalar", and the two give the same result:
\begin{align}
\omega(v) = v(\omega) \in \Bbb{F}
\end{align}
For question $2$, you're just plugging in a special case of $X=\frac{\partial}{\partial x^a}, Y=\frac{\partial}{\partial x^b}$, and $\omega = dx^m$. Then, starting from the first equation, we have:
\begin{align}
R\left(dx^m, Z^n\frac{\partial}{\partial x^n}, \frac{\partial}{\partial x^a}, \frac{\partial}{\partial x^b}\right) &= dx^m\left( \nabla_a \nabla_b Z - \nabla_b \nabla_a Z - \nabla_{\left[\frac{\partial}{\partial x^a}, \frac{\partial}{\partial x^b}\right]}Z\right)
\end{align}
Now, use multilinearity on both sides, and the definition of tensor indices: $T^{i_1,\dots, i_r}_{\qquad j_1, \dots, j_s} := T\left(dx^{i_1}, \dots, dx^{i_r}, \frac{\partial}{\partial x^{j_1}}, \dots, \frac{\partial}{\partial x^{j_s}}\right)$ (see lecture $3$) to get
\begin{align}
R^{m}_{\,\, nab}Z^n &= (\nabla_a\nabla_bZ)^m-(\nabla_b\nabla_aZ)^m - \left(\nabla_{\left[\frac{\partial}{\partial x^a}, \frac{\partial}{\partial x^b}\right]}Z\right)^m.
\end{align}
In general if you want to extract the equation in index form, just plug in the appropriate basis vector fields and covector fields.
