Why would this algorithm be vulnerable to a preimage attack?
Firstly, remember what is the pre-image attack*;


The pre-image attack is that given a hash value $h$ one need to find an $x$ such that $h = \operatorname{Hash}(x)$. The founded value doesn't need to be the pre-image that is used for the given hash value rather it can be any input value that its hash matches the given hash value, i.e. $x' \neq x$ and $h = \operatorname{Hash}(x')$.


Let call $\operatorname{H}_\ell (M) = (\ldots(((m_1 \wedge m_2) \vee  m_3) \wedge m_4) \vee m_5) \ldots [\vee|\wedge]\, m_{n-2})$ i.e words just before the last two operation. And

$$\operatorname{H}(M) = \big(\operatorname{H}_\ell (M)  [\vee|\wedge]\, m_{n-1}\big) [\vee|\wedge]\, m_n$$

The number of the blocks determines the order of the last two operations, the $\vee$ is the last or $\wedge$.

Now the trick is that


by using  $\wedge$ you can make any bit zero  $x \wedge 0 =0$ and 
by using $\vee$ you can make any bit one $x \vee 1 =1$.


Therefore only using the last two blocks - actually using only two blocks - the attacker can create any hash value.

There are much more simple attacks as a note by Fgriue:


$h = \operatorname{H}(h)$
$h = \operatorname{H}(h\mathbin\|h)$ since $h = h \wedge h $
$h = \operatorname{H}(1^n\mathbin\|h)$ since $h = 1^n \wedge h $
$h = \operatorname{H}(h\mathbin\|h\mathbin\|h)$ since $h = (h \wedge h) \vee h $
$h = \operatorname{H}(\bar h\mathbin\|h\mathbin\|h)$ since $h = (\bar h \wedge h) \vee h $




* The formal definition can be found in this seminal work Cryptographic Hash-Function Basics: Definitions, Implications, and Separations for Preimage Resistance, Second-Preimage Resistance, and Collision Resistance by P. Rogaway and T. Shrimpton.