I guess my answer will be in the category of complex manifolds, therefore we definitely need to assume that $X$ is smooth. I am not that familiar with this language, so I apologize for any stupid mistakes coming from this. 

Assume that $D$ is simple normal crossing and that $Y$ is smooth. Denote by $X_0 = X \setminus D$, $X_1 = D \setminus \text{Sing}(D)$, and inductively $X_{k+1}$ is the non-singular part of $\text{Sing}(\overline{X_k})$. Then the sections of $\mathcal{T}_X(- \log D)$ are those holomorphic vector fields on $X$, which are tangent to all strata $X_k$. I think a sufficient criterion for surjectivity is:

**Y intersects all the strata $X_k$ transversally (in the sense $T_p Y + T_p X_k = T_p X$ for all $p \in Y \cap X_k$)**

Indeed, to check whether the morphism $i^* \mathcal{T}_X(- \log D) \to \mathcal{N}_{Y/X}$ of locally free sheaves on $Y$ is surjective, it suffices to check in the fibres. At a point $p \in Y \cap X_k$ the divisor $D$ has equation $z_1 z_2 \ldots z_k =0$ in local coordinates and a local basis of $\mathcal{T}_X(- \log D)$ is given by 
$$z_1 \frac{\partial}{\partial z_1}, \ldots, z_k \frac{\partial}{\partial z_k}, \frac{\partial}{\partial z_{k+1}} \ldots, \frac{\partial}{\partial z_{\dim X}}.$$ 
The assumption, that $Y$ intersects $X_k$ transversally means that $T_p Y$ together with $\frac{\partial}{\partial z_{k+1}}|_p \ldots, \frac{\partial}{\partial z_{\dim X}}|_p$ span the vector space $T_p X$, thus the sections $\frac{\partial}{\partial z_{k+1}} \ldots, \frac{\partial}{\partial z_{\dim X}}$ restrict to generators of the fibre of $\mathcal{N}_{Y/X}$ at $p$. 

If we drop the assumption that $Y$ is smooth, we might be able to phrase the transversality condition in terms of the matrix of derivatives of a regular sequence $x_1, \ldots, x_l$ defining $\iota$ along tangent directions to $X_k$ having rank $l$. [For this see also the discussion in the comments.]