On the canonical isomorphism between $V$ and $V^{**}$

The first thing you mention is that, if you fix a basis for $V$, you get a non-canonical isomorphism $V\cong V^*$. Similarly, fixing a basis for $V^*$, you get a non-canonical isomorphism $V^*\cong (V^*)^*=V^{**}$. The "magic" is that, when you compose these two isomorphisms, you get an isomorphism $V\cong V^{**}$ which of course depends on your choice of bases, but which is exactly equal to the canonical isomorphism $$\iota: V\to V^{**};\quad \iota(v)=\iota_v,$$ regardless of what choices you made to define $\Phi$ and $\Phi^*$. The two extra equations you give show that $$\Phi^*\circ\Phi = \iota.$$ What may confuse you is that, in order to show these two maps are equal, we choose bases. But this is normal: we need these bases to define $\Phi$ and $\Phi^*$ to start with. But note that the two extra equations hold regardless of what bases were chosen.

The canonical map $\theta$ from a vector space to its double-dual has a particularly clean pointwise forumula:

$$\theta(v)(\omega) = \omega(v)$$

where $v$ is a vector and $\omega$ is a covector (an element of $V^*$).

Since $\omega$ ranges over all covectors, this gives a pointwise definition of $\theta(v)$. And since $v$ ranges over all vectors, this in turn gives a pointwise definition of $\theta$.

This notation is a little strange at first, but it's quite useful once you're used to it. $\theta(v)$ is an element of $V^{**}$, and so it makes sense to evaluate $\theta(v)$ at $\omega$ to get a scalar.