If we use the notation $(TM, p_M, M)$ for the tangent bundle of any manifold $M$, then you are right to think that $T^{\ k}M$ has $k$ natural vector bundle structures over $T^{\ k-1}M$ and so on down to $M$, making a diagram which is a $k$-dimensional cube. Such a structure is a $k$-fold vector bundle (See articles by Kirill Mackenzie) and $T^{\ k} M$ is a particularly symmetrical one. An easy way of writing the $k$ bundle maps would be
$$T^{\ a}(p_{T^{\ b} M}) $$
for $a+b+1 = k$, where this means that we are taking the $a$-th derivative of the tangent bundle projection $T\ (T^{\ b}M)\to T^{\ b}M$, yielding a map now from $T^{\ k}$ to $T^{\ k-1}$.
To see the symmetries of $T^{\ k} M$ it is more convenient to describe the functor in a direct way rather than as a $k$-fold composition -- just as you might think of tangent vectors as infinitesimal curves, you can think of points in $T^{\ k} M$ as infinitesimal maps of a unit $k$-cube into $M$. The restrictions to the $k$ faces of the cube (going through the origin) gives your $k$ maps.
From that point of view, it is clear that you can permute the $k$ coordinate axes and get another map of a cube into $M$, so that the functor $T^{\ k} M$ has a $S_k$ group of natural-automorphisms.
Incidentally, to make the above into a definition of $T^{\ k} $, you could do the following:
Consider the fat point $fp$, which you should think of as a space whose smooth functions form the ring $\mathbb{R}[x]/(x^2)$. Then the tangent bundle $TM$ can be thought of as the space of maps from the fat point to $M$, i.e. $TM=C^\infty(fp,M)$. Such maps, by the way, are just algebra homomorphisms from the algebra $C^\infty(M,R)$ to $\mathbb{R}[x]/(x^2)$. You can check that such a map has two components $f_0 + f_1 x$, and that $f_0$ is a homomorphism to $\mathbb{R}$ defining a maximal ideal (i.e. a point $p$ in $M$) and $f_1$ defines a derivation (i.e. a vector at $p$).
In precisely the same way you can consider a cubical fat $k$-point $kfp$ with functions
$$\mathbb{R}[x_1,...,x_k]/(x_1^2,...,x_k^2)$$
and then define $T^{\ k} M = C^\infty(kfp,M)$. Then you can see the symmetries as automorphisms of the above algebra, and the $k$ maps you ask about as homomorphisms $C^\infty(kfp)\to C^\infty\big((k-1)fp\big).$
There are probably more subtle things to be said about these higher iterated bundles but I hope the above is at least correct.
Let me give an example showing that the normality hypothesis is necessary.
Let $Y=\mathbb{P}^1$ with natural $G=\mathbb{G}_m$-action. Let $X$ be the $G$-variety obtained by glueing transversally the two fixed points $0$ and $\infty$. Consider the line bundle $\mathcal{O}(l)$ with $l\neq 0$ on $Y$ and glue the fibers over $0$ and $\infty$ using any linear isomorphism to obtain a line bundle $\mathcal{L}$ on $X$. Suppose that $\mathcal{L}$ has a $G$-linearisation. Pulling it back to $Y$, we obtain a $G$-linearisation of $\mathcal{O}(l)$ on $Y$ such that $G$ acts on the fibers over $0$ and $\infty$ with the same character. However, the description of the $G$-linearisations of $\mathcal{O}(l)$ when $l\neq 0$ shows that this is not possible (more precisely, for any $G$-linearisation of $\mathcal{O}(l)$, the characters through which $G$ acts on the fibers over $0$ and $\infty$ differ by the character $t\mapsto t^l$). This argument shows moreover that no multiple of $\mathcal{L}$ has a $G$-linearisation.
Note that it follows that there is no ample $G$-linearised line bundle on $X$.
As for the second question, the natural map to study is more likely to be $Pic^G(X)\to Pic(X)^G$ where $Pic(X)^G$ denotes the group of line bundles whose class in $Pic(X)$ is $G$-invariant. When $X$ is normal and proper, its Picard group is an extension of a discrete group by an abelian variety so that if $G$ is linear connected, $G$ acts necessarily trivially on $Pic(X)$ and $Pic(X)^G=Pic(X)$. However, when $X$ is not normal, this is not the case anymore (for instance in the above example).
Moreover, the arguments in Dolgachev's notes extend to show that, if $X$ is an integral proper variety over an algebraically closed field endowed with an action of a connected linear algebraic group $G$, there is an exact sequence : $$0\to K\to Pic^G(X)\to Pic(X)^G\to Pic(G).$$
A more general statement, that does not assume $X$ proper or $G$ affine, may be found in [Brion, On linearization of line bundles, arXiv:1312.6267, Proposition 2.10].
Best Answer
If $L$ is any line bundle over a compex manifold $X$, a square root of $L$ is a line bundle $M$ such that $M^{\otimes2}=L$. So your guess in part (2) is correct.
This square root (if it exists) is not unique in general, and two of them will differ by a $2$-torsion line bundle, that is a line bundle $\eta$ such that $\eta^{\otimes 2}$ is trivial. In particular, if $\textrm{Pic}(X)$ is torsion free, then there is at most one square root.
In some cases no square root exists. Some general results are:
A line bundle of degree $0$ has always at least one square root. This because $\textrm{Pic}^0(X)$ is a complex torus, hence a divisible group (in fact, there are roots of any order).
A line bundle over a Riemann surface of genus $g$ has a square root if and only if it has even degree. The number of different square roots equals in this case $2^{2g}$, the number of $2$-torsion points in $\textrm{Pic}^0(X) \cong \textrm{Jac}(X)$.
If $L$ is effective, that is $H^0(X, L) \neq 0$, and $Z \subset X$ is the zero locus of a holomorphic section of $L$, then the existence of a square root of $L$ is equivalent to the existence of a double cover $Y \to X$ branched over $Z$. In particular, non-trivial square roots of the trivial bundle correspond to non-trivial unramified double covers of $X$.
The square root of the canonical bundle of the Riemann Sphere $S$ is unique, since $\textrm{Pic}(S)=\mathbb{Z}$, and it is isomorphic to $\mathcal{O}(-1)$, the dual of the hyperplane bundle (the unique line bundle of degree $1$, whose transition function is $z \to 1/z$).
A readable introduction to spinor bundles is provided in the book of MOORE "Lectures on Seiberg-Witten invariants".