Pullback of invertible sheaves corresponds to pull-back of line bundles in the geometric sense.
In usual topology, a (complex) line bundle is a family of one-dimensional $\mathbb C$-vector spaces varying over a topological space $X$. More precisely, it is a topological space $V$, equipped with a continuous map $V \to X$, a zero section $X \to V$,
a scalar multiplication action $\mathbb C \times V \to V$ compatible with the projections to $X$, an addition map $V \times_X V \to V$, again compatible with the projections to $X$, all satisfying some evident axioms, which more or less amount to requiring that these structure endow each fibre $V_x$ of $V$ over a point $x$ of $X$ with the structure of a one-dimensional $\mathbb C$-vector space, in such a way that $V$ is locally trivial.
Now if $\varphi: Y \to X$ is a continuous map, we may form the fibre product $\varphi^*V := V\times_X Y,$ and this is a line bundle over $Y$.
If you just think about how fibre product is defined, you see that the fibre of $\varphi^*V$ over a point $y \in Y$ is canonically identified with the fibre of $V$ over $\varphi(y)$. So if we think of $V$ as being the family $V_x$ of lines parameterized by the points $x \in X$, then $\varphi^*V$ is the family of lines $V_{\phi(y)}$ parameterized by $y \in Y$.
We may form the sheaf of sections $\mathcal L$ of $V$; this is a locally free sheaf of rank one over the sheaf of continuous $\mathbb C$-valued functions on $X$. The sheaf of sections of $\varphi^*V$ is then $\varphi^*\mathcal L$.
There is an exercise somewhere in Hartshorne which describes the analogue of the
above theory of vector bundles in the algebro-geometric setting; everything
goes over more-or-less the same way. I am now going to work in the algebro-geometric setting.
It is not so easy to relate the sections of $V$ to those of $\varphi^*V$;
it depends very much on the space $Y$ and the morphism $\varphi$. For example,
if $Y$ is just a point $x \in X$, then $\varphi^*V$ is just the line $V_x$, and its space of
sections is one-dimensional (indepdently of what $V$ is). But it might be
that $V$ has no sections that are non-zero at $x$.
More generally, a non-trivial bundle on $X$ might pull-back to something trivial
on $Y$.
So you have to analyze this question on a case-by-case basis. Cohomology is one of the basic tools available here, but I am guessing that you're not at the point of using cohomology yet.
E.g. if $Y = \mathbb P^1$ and $\varphi: Y \to \mathbb P^2$ embeds $Y$ as a plane conic, then this image is a degree two curve, so $\mathcal O(1)$ pulls back to a degree two sheaf on $Y$, i.e. to $\mathcal O(2)$. Thus its space of global
sections is three dimensional. So in this case $\varphi^*$ induces an isomorphism on spaces of global sections.
E.g. If $Y = \mathbb P^1$ but $\varphi: Y \to \mathbb P^2$ embeds it as a line,
then $\varphi^*\mathcal O(1)$ is just $\mathcal O(1)$ on $\mathbb P^1$ (a line is a degree one curve), and so the map on sections is surjective, with a one-dimensional kernel (which is precisely the linear form cutting out the image
of $\varphi$).
As Martin Brandenburg indicates in an answer, the sections $s_i$ give a projective embedding in the following concrete way: locally we may trivialize $V$, and so regard the $s_i$ as simply functions; we thus obtain a morphism
$x \mapsto [s_0(x): \cdots : s_n(x)].$ Note that if we change trivialization,
then the interpretation of the $s_i$ as functions changes by a nowhere zero function (the same function for all the $s_i$), which means that the point
in $\mathbb P^n$ doesn't change. Thus the map is well-defined independent
of the trivialization.
Finally, it might be helpful to know that in topology, the complex line bundles
are classified (up to isomorphism) by homotopy classes of maps to $\mathbb C P^{\infty}$. Again, the map is given by pulling back $\mathcal O(1)$.
The situation in algebraic geometry is analogous to this, but more rigid: topological isomorphism is much less rigid than algebraic isomorphism (hence only the homotopy class of $\varphi$ matters), and all line bundles have lots of sections. In algebraic geometry, only sufficiently positive (i.e. sufficiently ample) bundles arise from maps to projective space.
I don't believe you can find a totally coordinate-free way to phrase things, unless you're just tautologically hiding the coordinates. The choice of $n+1$ global sections of $\mathcal{L}$ which generate are the coordinates on $\Bbb{P}^n_S$, so you're implicitly starting with a set of coordinates. Moreover, the line bundle $\mathcal{L}$ does not uniquely determine a collection of global sections which give a unique morphism $T\to\Bbb{P}^n_S$ -- this choice is essential, and can give you different morphisms.
Of course, you can rephrase your starting assumptions. Let $f : T\to S$ be the original morphism. Your sections are equivalent to the data of a surjection $f^\ast\mathcal{O}_S^{n+1} \cong\mathcal{O}_T^{n+1}\to\mathcal{L},$ and such a surjection is precisely the data of an $S$-morphism $T\to\Bbb{P}(\mathcal{O}_S^{n+1}) \cong \Bbb{P}^n_S,$ as if $\mathcal{E}$ is a locally free sheaf (of finite rank) on $S,$ then $$\operatorname{Hom}_S(T,\Bbb{P}(\mathcal{E}))\cong\{\textrm{invertible quotients of }f^\ast\mathcal{E}\}.$$
But there are many morphisms $T\to\Bbb{P}^n_S$ in general, and they will depend on the coordinates you pick, which your example does not seem to.
As I described in my comment, if we don't require $n$ such that $n+1$ is the minimum number of global sections required to generate $\mathcal{L}$, then there really isn't a nice "coordinate-free" way to describe the morphism. Let $T = S = \operatorname{Spec}k,$ and let $\mathcal{L} = \mathcal{O}_T,$ and let $n=1.$ Then the choice of two global sections is the choice of a point $[a_0 : a_1]\in\Bbb{P}_k^1,$ but this choice will give you completely different points as you change what global sections you pick. You could say that this morphism simply picks out a point in $\Bbb{P}(\mathcal{E}),$ where $\mathcal{E}$ is free of rank $2,$ but you already knew this because $T = \operatorname{Spec}k.$
Even if we force $n$ to be such that $n+1$ is the minimum number of global sections required to generate $\mathcal{L},$ we cannot say too much. let $T =\Bbb{P}^1_k$, $S = \operatorname{Spec}k,$ and let $\mathcal{L} = \mathcal{O}(1).$ Then we need $n+1$ global sections to generate $\mathcal{L},$ but we have many choices of what these may be. We might choose $x_0,\dots, x_n,$ but we might also choose $x_0 + x_1 + \dots + x_n, x_1 + \dots + x_n,\dots, x_{n-1} + x_n, x_n.$ The corresponding maps are distinct maps to projective space, as the two invertible quotients are not equivalent: there is no isomorphism $i : \mathcal{L}\to\mathcal{L}$ making the diagram
$$\require{AMScd}
\begin{CD}
\mathcal{O}_T^{n+1} @>(x_0,x_1,\dots,x_n)>> \mathcal{L} \\
@V{\operatorname{id}}VV @VViV\\
\mathcal{O}_T^{n+1} @>>(x_0 + \dots + x_n, \dots, x_n)> \mathcal{L}
\end{CD}
$$
commute (recall that $\operatorname{Isom}(\mathcal{L},\mathcal{L})\cong\mathcal{O}(T)^\times$ for any line bundle $\mathcal{L}$ on $T$). That is, there is no isomorphism of invertible quotients between $(x_0,\dots, x_n) : \mathcal{O}_T^{n+1}\to\mathcal{L}$ and $(x_0 + \dots + x_n,\dots, x_n) : \mathcal{O}_T^{n+1}\to\mathcal{L}.$
Perhaps you have something particular in mind when you say coordinate-free, or a particular type of description you're looking for. If this is the case and my answer does not address those, it would be helpful to narrow down exactly what you're looking for and requiring, and add that to the question.
Best Answer
These two descriptions are equivalent. It suffices to show that a line bundle is generated by a set of global sections if and only if they have no common zeroes. For a set of global sections $(s_i)_{i \in I}$ of a sheaf $\mathcal F$ of $\mathcal O_X$-modules, the surjectivity of the induced morphism of sheaves $\bigoplus_{I}\mathcal O_X \to \mathcal F$ may be checked on the level of stalks. But for a line bundle, the stalk $\mathcal F_x$ is just isomorphic to $\mathcal O_{X,x}$, so surjectivity on stalks amounts to not vanishing simultaneously.