An Erdős-Szekeres-type question

The answer is YES if all the $d_i$ are at least $\lfloor d/2\rfloor$, and NO otherwise.

For the NO: as the cyclic polytopes $C_d(n)$ are $\lfloor d/2\rfloor$-neighborly, there are arbitrarily large point sets where no simplex spanned by $\lfloor d/2\rfloor$ points intersects any simplex (even $d$-simplex) spanned by other points.

For the YES: We rely on the high-dimensional version of Erdös-Szekeres as quoted by Gil Kalai.
(For references, this is Exercise 6(i) of Section 7.3, page 126 of Grünbaum's polytopes book, where however only "neighborly" is claimed/stated. The full claim for "cyclic" with a proof is given in Proposition 9.4.7, page 398, of the five-author oriented matroid by Björner et al. Indeed we get - and will use - the stronger condition that also all subpolytopes will be cyclic.)

Thus we can assume that our point set contains/is the vertex set of of a $d$-dimensional cyclic polytope on $(d_1+1)+\cdots+(d_k+1)$ vertices. Also assume w.l.o.g. that $d_1\ge d_2\ge\cdots\ge d_k$. Now if we assign the vertices to simplices by taking rounds - that is, first vertex to $\Delta_1$, second to $\Delta_2$, ... , $k$-th to $\Delta_k$, $(k+1)$-st to $\Delta_1$ again etc., we get a partition of the vertex set into simplices. One checks that they pairwise intersect using the combinatorics of cyclic polytopes and the Gale evenness criterion. For this, the key fact is that when we look at the vertices of $\Delta_i$ and $\Delta_j$, then they come in alternating order on the cyclic polytope. Together, they have at least $d+2$ vertices. Finally, the combinatorics of a cyclic polytope on $d+2$ vertices is that of two simplices that intersect in point that is relative-interior to both of them.

The answer to the first question is yes (with $N = 7$). Consider a $2$-dimensional simplicial complex $K$ with 7 vertices and with all possible triangles. Assume there are 7 points in $\mathbb R^4$. You can construct a linear mapping $f \colon |K| \rightarrow \mathbb R^4$ ($|K|$ denotes the geometric realization of the complex). It is well known that $K$ does not embed into $\mathbb R^4$. More precisely, so called Van Kampen obstruction of this complex is nonzero, and this implies that for any continuous map $f'\colon |K| \rightarrow \mathbb R^4$ there are two vertex disjoint simplices $\sigma, \tau$ of $K$ whose images $f(|\sigma|), f(|\tau|)$ intersect. If you aply this result on $f$ you get the desired conclusion. In an unlikely case that $\sigma$ or $\tau$ have smaller dimension then 2, you simply extend them to triangles (this may occur only if the points are not in generic position).

This reasoning extends to the case $k = 2$, $d_1, d_2 = m$ for some parameter $m$, $d = 2m$ and $N = 2m + 3$. (At the moment, I am not able to answer the general question, however.)

It is useful to know the following Erdos-Szekeres fact: For every k > d there is N(k,d) so that every N points in general position in R^d contains k points in "cyclic position". We say that d points $x_1, x_2,...x_d$ are in cyclic position if all the simplices $x_{i_1},...,x_{i_{d+1}}$ have the same orientation. This fact follows from Ramsey's theorem. (With very large N(k,d).) It implies various results of the kind ask here if they refer to properties of points in cyclic positions.

This gives a complete answer for the case $k=2$ of the original question. Indeed it is useful to think about the original problem as for which sizes $d_1,d_2\dots,d_k$ whose sum is $(d+1)(k-1)+1$ if N is large enough and we have $N$ points in $R^d$ we can find a subset of $(d+1)(k-1)+1$ points with Tverberg partition of sizes $d_1,d_2,\dots,d_k$. When it comes to Radon partitions points in cyclic position are "cannonical". For larger values of $k$ I dont know the precise situation. We can look at lexicographic sequence of points on the moment curve and this is a property inherited by subsequences. (So it will exclude plenty of $d_i$ sequences.) But I am not sure every large set of points "contains" a lexicographic sequence of points on the moment curve. ("contains" in terms of having equivalent Tverberg's behavior.) So there is more to explore.

By the way, an interesting higher dimensional question is what is the number f(n,d) so that every f(n,d) points in general position in $R^d$ contains $n$ points in convex position. This is monotonic non-increasing in $d$.