Why do resorcinol and iodine react to give 2-iodobenzen-1,3-diol?
Reading this question, I realized that OP is very new to organic chemistry, and in need for learning a lot about electrophilic aromatic substitution reactions. Thus, I recommend that OP should concentrate on the electrophilic aromatic substitution reactions and read the chapters of OP's textbook dedicated to that subject. Said that, I'm going to answer plethora of subjective questions, which should be answered and cleared to show the way to right direction.

I got told by my lab assistant that the mechanism is an electrophilic aromatic substitution in which the Resorcinol (Benzin-1,3-diol) acts as the electrophile and Iodine as nucleophile.

First, resorcinol is not called benzin-1,3-diol (may be in an old book). It's correct name is 1,3-dihydroxybenzene. It is true that the reaction is an electrophilic aromatic substitution. However, the electrophile here is polarizable $\ce{I2}$ molecule because it exsists in $\ce{I^{\delta +}-I^{\delta -}}$ form under reaction conditions (where $\ce{I^{\delta +}}$, which lacks electrons is the reactive center). Aromatic nucleus with two hydroxy group is electron rich (hydroxy groups activate the ring), hence acts as the nucleophile.

But isn't the +M Effect stronger?

Yes, mesomeric effect (old resonance effect) is stronger than inductive effect. Here you have two hydroxy groups to give strong $(+M)$-effect to activate the aromatic nucleus. Hence the reaction is o,p-directive. As a result, you can expect either 2-iodo-1,3-dihyroxybenzene or 4-iodo-1,3-dihyroxybenzene or 2,4-diiodo-1,3-dihyroxybenzene or etc. as your product or mixture of them based on the reaction condition used (Ref.1 &amp; 2).

Why is the Product 2-Iodo-1,3-diol and not 4-Iodobenzen-1,3-diol?

As I explained before, the reaction conditions govern the product ratio (see the PDF of Ref.1 I attached below for better understanding). The iodination of resorcinol is not a straight forward reaction. Slight changes of reaction conditions gives you different product. For example, the major product given in the reaction using $\ce{I2/NaHCO3}/\pu{0 ^\circ C}$ condition is 2-iodo-1,3-dihydroxybenzene as indicated in the OP's question. In a large scale reaction, the yield obtained was only 66% (Ref.3). Using the same conditions, Ref.1 reported 77% isolated yield. Keep in mind that, some of 2-iodo-1,3-dihydroxybenzene isomerized to 4-iodo-1,3-dihydroxybenzene (major) and 4,6-diiodo-1,3-dihydroxybenzene in an acidic medium at high temperature (Ref.1). Thus, temperature plays an important role in percent yield in the reaction.
On the other hand, when the electrophile ($\ce{I+}$ source is changed to $\ce{ICl}$, the main product is 4-iodo-1,3-dihydroxybenzene (4-iodoresorcinol) with 4,6-diiodo-1,3-dihydroxybenzene (4,6-diiodoresorcinol) as minor product (Ref. 1 &amp; 2). Ref.2 also stated that use of nascent iodine $(\ce{KIO3/KI/HCl})$ has resulted in the regiospecific 2,4-diiodonation of resorcinol at room temperature (which is a rapid reaction). The $\ce{^1H}$-NMR spectrum of this reaction mixture has revealed only two products, 2,4-diiodoresorcinol as major product (56% isolated yield) and 2,4,6-triiodoresorcinol as minor product in $89:11$ ratio.
References:

Ib Thomsen, and Kurt B. G. Torssell, &quot;Iodination of Resorcinol, 5-Methoxyresorcinol, Phloroglucinol and Resorcyclic Acid,&quot; Acta Chemica Scandinavica 1991, 45, 539-542 (DOI: 10.3891/acta.chem.scand.45-0539)(PDF).
Frederick L. Weitl, &quot;Regiospecific 2,4-diiodination of resorcinol with nascent iodine,&quot; J. Org. Chem. 1976, 41(11), 2044–2045 (DOI: https://doi.org/10.1021/jo00873a040).
Shin-ichiro Tsujiyama and Keisuke Suzuki (submitted); David B. Guthrie, Heather M. Gibney, and Dennis P. Curran (checked), &quot;Preparation of Benzocyclobutenone Derivatives Based on an Efficient Ggeneration of Benzynes [Bicyclo[4.2.0]octa-1,3,5-trien-7-one, 5-(phenylmethoxy)-],&quot; Org. Synth. 2007, 84, 272-284; Coll. Vol. 11 2009, p. 488-497 (DOI: 10.15227/orgsyn.084.0272).