I'll write $H$ for the Hilbert scheme and $[Y]$ for the point of $H$ corresponding to a subscheme $Y \subseteq X$. When $Y$ and $X$ are both smooth, the claim is that the tangent space to $H$ at $[Y]$ is the global sections of $(T_X)|_Y/T_Y$.

**Intuition 1** (This is basically Scott's comment.) Given a holomorphic vector field $\theta$ on $X$ near $Y$, we can flow $Y$ along $\theta$ to get a family $Y(t)$ of complex submanifolds of $X$, and thus a path $[Y(t)]$ in $H$. It is plausible (and true!) that the first order variation $\frac{d}{dt} [Y(t)]$, which is a tangent vector to $H$, only depends on $\theta|_Y$. Moreover, if $\theta$ is tangent to $Y$, then $Y$ flows to itself, so it makes sense that only the image of $\theta$ in the quotient $T_X/T_Y$ matters.

**Intuition 2** Let's go the other way: Let $[Y(t)]$ be a path through $H$ with $Y(0) = Y$. For $t$ in some open disc $D$ around $0$, we can smoothly (not holomorphically!) trivialize the family and thus get a map $\phi: Y \times D \to X$. For every point $y \in Y$, we have a path $\phi(y,t)$ in $X$, whose derivative at $0$ lies in $T_X(y)$. But changing the trivialization can change this derivative by vectors in $T_Y(y)$, so the intrinsic thing is a vector in $T_X(y)/T_Y(y)$ for each $y \in Y$ -- in other words, a section of the normal bundle.

**Bonus** We have a long exact sequence $0 \to T_Y \to (T_X)|Y \to N_{Y/X} \to 0$ of vector bundles on $Y$. So we have a long exact sequence $H^0(Y, T_X|Y) \to H^0(Y, N_{Y/X}) \to H^1(Y, T_Y)$. The final vector space, $H^1(Y, T_Y)$, describes deformations of the complex structure of $Y$ and, indeed, the map $H^0(Y, N_{Y/X}) \to H^1(Y, T_Y)$ says, when you wiggle $Y$ inside $X$, how the complex structure on $Y$ changes. The tangent vectors coming from $H^0(Y, T_X|Y)$ are the ones that don't change complex structure. Intuitively, if we have an actual section of $T_X$, then we can flow the embedding $Y \to X$ along it to get a holomorphic map $Y \times D \to X$.

Of course, all of this is intuition only in the case that $X$ and $Y$ are smooth. In order to see that $\mathrm{Hom}(I_Y/I_Y^2, \mathcal{O}_Y)$ is the version which is still right when you put singularities in, I recommend Count Dracula's method.