If $\mathcal{F}$ is a foliation on an $n$-manifold $M$, then the tangent bundle to $\mathcal{F}$, which I'll call $\mathrm{T}\mathcal{F}$, is a $k$-plane distribution on $M$, where $k$ is the dimension of the leaves of $\mathcal{F}$. This is a standard fact about foliations but I have been unable to find an explanation why $\mathrm T \mathcal F $ is a $k$-plane distribution (either one of my own or from some source).
That is, given $\mathcal{F}$ I can construct a subset of $\mathrm T M$ in the following manner. For each $x \in M$ there is a leaf $L_x \in \mathcal F$ that passes through $x$. Since $L_x$ is an injectively immersed submanifold, its tangent bundle is a subbundle of $\mathrm T M$, so I define $\mathrm T_x \mathcal F = \mathrm T_x L_x$. Why is $\mathrm T \mathcal F = \bigcup_{x \in M} \mathrm T_x \mathcal F$ actually a vector bundle on $M$?
I'm not entirely sure what your question is.
Correct me if I'm wrong, but it seems that you understand how we associate a $k$-plane of the tangent space at every point to each point. Is what you are confused about then why this subset of the tangent bundle is itself a vector bundle?
For that, all we need to check is that it is locally trivializable. However, this is immediate from considering a foliated coordinate neighborhood of a point $\phi: U\to V\subset \newcommand{\RR}{\Bbb{R}}\RR^k\times\RR^{n-k}$ (with $V$ a rectangular neighborhood). Then in this foliated coordinate neighborhood, at every point of $V\cap \RR^p$, the tangent space is just $\RR^k\subset \RR^n$ (identifying the tangent space of points in $\RR^n$ with $\RR^n$ in the usual way). Then a basis of sections for the tangent bundle of the plaques in this neighborhood is given by $\mathbf{e}_i:x\mapsto e_{i,x}$, for $1\le i\le k$ where $e_{i,x}$ is the $i$th standard basis vector regarded as an element of the tangent space to $x$. Thus the tangent bundle of the foliation is locally trivializable, hence itself a vector bundle.