This is a nice question which permits to illustrate the superiority of schemes over classical varieties.
Every ideal $I\subset k[x_1,...,x_n]$ yields a subscheme $V(I)\subset \mathbb A^n_k$ and this yields is a perfect bijective correspondence between ideals of $k[x_1,...,x_n]$ and closed subschemes of $\mathbb A^n_k$.
Consider for example the ideals $I=(x,y),J=(y,x^2), K= (x^2,y^2) \subset k[x,y]$.
They correspond to three schemes $V(I), V(J), V(K)\subset \mathbb A^2_k$ which are different although their underlying set is the same singleton set $\lbrace (0,0) \rbrace$.
The best proof that they are different is that their tangent spaces at their single point are the different subvectorspaces of $k^2$ equal respectively to zero, the $x$-axis $y=0$ and the whole space $k^2$.
Classically one would be baffled because one would have to say that $I,J,K$ define the same subvariety of $\mathbb A^2_k$ but that the ideals $J$ and $K$ don't allow to define the tangent space.
The way out would have been to declare that the correct ideal of a subvariety $V$ is the ideal $i(V)$ of polynomials vanishing on the variety $V$, which corresponds (by the Nullstellensatz) in the case of an algebraically closed field to $V(\sqrt I)$ where $I$ is any ideal for which $V(I)=V$ set-theoretically .
The scheme point of view is clearer, more refined and perfectly solves the conundrum posed in the question about the tangent space of a variety defined by two different sets of polynomials $f_1,\dots f_r$ and $g_1,\dots g_s$: we do not have $T_p=T'_p$ in general because we do not have $V(f_1,\dots f_r)=V(g_1,\dots g_s)$ scheme-theoretically