Given a linear map $g \colon V \rightarrow W$ between finite dimensional vector spaces, we can form the exact sequence
$$ 0 \rightarrow V \xrightarrow{\Delta} V \times_{W} V \rightarrow (V \times_{W} V )/ \operatorname{im}(\Delta) \rightarrow 0 $$
where $V \times_{W} V = \{ (v,v') \, | \, g(v) = g(v') \}$ and $\Delta(v) = (v,v)$ is the diagonal embedding. Then we can see that $(V \times_{W} V) / \operatorname{im}(\Delta)$ is canonically isomorphic to $\ker g \subseteq V$ via the subtraction isomorphism $[(v,v')] \mapsto v - v'$ (whose inverse is given by $v \mapsto [(v,0)]$).
This applies fiberwise to the situation you describe for manifolds. Let's assume that $f$ is a surjective submersion (this is enough to guarantee that the fibered product actually exists in the category of smooth manifolds). Fix $x \in X$ and pull back the short exact sequence via $\Delta$ to obtain
$$ 0 \rightarrow \Delta^{*} \left( T(\operatorname{im} \Delta) \right)|_{x}
\rightarrow \Delta^{*} T(X \times_{Y} Z)|_{x} \rightarrow \Delta^{*} (N_{\operatorname{im}{\Delta} \hookrightarrow X \times_{Y} X})|_{x} \rightarrow 0 $$
Using the identifications
$$
T(X \times_{Y} X)|_{(x,x')} = \{ (v,v') \in T_x X \times T_{x'} X, \, | \, df|_{x}(v) = df|_{x'}(v') \}, \\
\Delta^{*} T(X \times_{Y} Z)|_{x} = T_x X \times_{T_{f(x)} Y} T_x X, \\
\Delta^{*} \left( T(\operatorname{im} \Delta) \right)|_{x} = \{ (v,v) \in T_x X \times T_x X \} \cong T_x X, \\
T_{X/Y}|_{x} = \ker \left( df|_{x} \right)
$$
we see that we are in the algebraic situation described in the beginning of the answer (with $V = T_x X, W = T_{f(x)} Y$ and $g = df|_{x}$).
To see what is going on geometrically, it might be useful to start with the case where $Y = \textrm{pt}$ so that the fibered product is just a regular product $X \times X$ and the relative tangent bundle is the regular tangent bundle of $X$. Take $X = \mathbb{R}$ and identify each equivalence class in a fiber of the normal bundle of $X$ inside $X \times X$ with a slanted line. The isomorphism then identities each such slanted line with its intersection with the (say) $x$ axis (which is the "unslanted, regular" tangent space).
You are correct that you may interpret $\Omega^1_{X/k}$ as $\Omega^1_{k[X]/k}$ and $\mathcal{O}_X$ as $k[X]$ if you want to avoid speaking of sheaves. Now, to answer your questions:
$\Omega^1_{X/k}$ modulo $m_P$: Let $A$ be a commutative ring, let $\mathfrak{m}$ be an ideal of $A$, and let $M$ be an $A$-module. Then $\mathfrak{m}M := \{am\mid a\in\mathfrak{m}, m\in M\}$ is an $A$-submodule of $M,$ and we may form the quotient module $M/\mathfrak{m}M,$ which is naturally a module over $A/\mathfrak{m}.$ As a side note, $M/\mathfrak{m}M$ is isomorphic to $M\otimes_A A/\mathfrak{m}$.
Dimension of $X$: The dimension of a variety or scheme is its dimension as a topological space. This means that the dimension of $X$ is the supremum of the lengths of chains of irreducible closed subsets of $X$:
$$\dim X := \sup_{n}\{Z_0\subsetneq Z_1\subsetneq \cdots\subsetneq Z_n\subseteq X\mid Z_i\textrm{ closed irreducible subspace of }X\}.$$
In an affine scheme $X = \operatorname{Spec}A,$ such a chain corresponds to a chain of prime ideals in $A,$ so that the dimension of $X$ is the same as the Krull dimension of $A.$
Locally free: This one is a little harder to make sense of without sheaves. You should know that to any morphism of schemes $f : X\to S,$ we may associate a sheaf of $\mathcal{O}_X$-modules $\Omega^1_{X/S}$ which is obtained by gluing together the sheaves of Kähler differentials on affine pieces. A sheaf $\mathcal{F}$ on $X$ is called locally free if for every point $x\in X,$ there exists an open neighborhood $U\ni x$ such that $\left.\mathcal{F}\right|_U\cong\mathcal{O}_U^n$ for some $n.$
Translating this back into a condition on your module, to be locally free in a neighborhood of $P$ means that there exists some $f\in A\setminus m_P$ such that the localization $M_f = M[f^{-1}]$ is isomorphic to a free $A_f = A[f^{-1}]$-module of some rank.
Best Answer
Let's do your example (1).
There is a short exact sequence $$0\to A\to A\mathrm dx\oplus A\mathrm dy\to\Omega^1\to 0$$ in which the first map maps $1$ to $2x\,\mathrm dx+3y\,\mathrm dy$. It follows that $\Omega^1$ is isomorphic to the module $$\frac{A\oplus A}{(2x,3y)}.$$
If the characteristic is $2$ or $3$, this looks like $A\oplus A/(y)$ or $A\oplus A/(y)$. If the characteristic is not one of those two, then it is clear, from this description, that if you localize at a prime not containing $x$ or $y$, then you get a free module. Now localize at one of the other ideals, and see what you get!