1- Your natural transformation can be seen as a functor $C \to D^{\Delta^1}$, which therefore induces a commutative square of $\infty$-categories

$\require{AMScd} \begin{CD} C_{x/} @>>> D^{\Delta^1}_{\alpha_x/} \\
@VVV @VVV \\
C @>>> D^{\Delta^1}\end{CD}$

and thus a morphism of fibers over $y\in C$. The fiber of the leftmost vertical map is $map_C(x,y)$ , and the fiber of the rightmost vertical map over the image of $y$, i.e. $\alpha_y$ sits in a (cartesian) square

$\require{AMScd} \begin{CD} map(\alpha_x,\alpha_y)@>>> map(fx,fy) \\
@VVV @VVV \\
map(gx,gy) @>>> map(gx,fy) \end{CD}$

For your purposes, you don't even need to know the square is cartesian, and then it just comes from the fact that $\Delta^1\times\Delta^1$ *is* a commutative square.

Here's maybe more detail : consider an arrow $g:x_1\to y_1$ in an $\infty$-category $E$, then $Fun(\Delta^2,E)\times_{Fun(\Delta^1,E)} \{g\}\simeq E_{/g}$, where we look at evaluation at $\Delta^{\{1,2\}}$.

Also recall that the forgetful map $E_{/g}\to E_{/x_1}$ is an equivalence (because the space of fillings for a given inner horn is contractible).

Now consider the following sequence of inclusions $\Delta^1\to \Delta^2 \to\Delta^1\times \Delta^1$ : the first map is inclusion at $\Delta^{\{0,1\}}$,and the second one is the diagonal arrow.

This provides you with functors $Fun(\Delta^1\times\Delta^1,E)\to Fun(\Delta^2,E)\to Fun(\Delta^1,E)$, where the second one induces an equivalence upon taking the fiber over $g$ and $x_1$ respectively. So you have functors $Fun(\Delta^1\times\Delta^1,E)\times_{Fun(\Delta^1, E)} \{g\}\to Fun(\Delta^2,E)\times_{Fun(\Delta^1,E)}\{g\}\to Fun(\Delta^1,E)\times_E \{x_1\}$, where the second one is an equivalence.

Now do the same for the other nondegenerate $2$-simplex in $\Delta^1\times\Delta^1$, where this time you'll look at another arrow $f: x_0\to y_0$ in $E$. This gives you a big diagram as follows,witnessing the fact that $\Delta^1\times\Delta^1$ is a commutative square:

(because AMScd does not support diagonal arrows, this is hard to draw on MO, so I just drew it and took a picture)

It all commutes, and the "wrong way" morphisms become invertible upon taking the appropriate fibers. On fibers (I'll let you figure out which fibers I mean), this then gives you the following, still commutative diagram :

which induces the desired commutative diagram, relating $map(f,g), map(x_0,x_1), map(y_0,y_1)$ and $map(x_0,y_1)$.

For your question
2-, the answer is not as easy because it depends on what your original definition of adjunction is. If you're just following *Higher topos theory*, then your statement is essentially 5.2.2.12 in that book, and the proof requires a number of preliminaries. If that's not what your definition is, you're going to need to be more precise about what you mean.