Yes, for both questions.

Q2: Since it's a local question, it suffices to consider $M$ open in $\mathbb{C}^n$ and $\mathcal{E},\mathcal{F}$ one dimensional trivial complex bundles. So let $\mathcal{O}$ denote the ring of holomorphic functions on $M$, let $z_1,\ldots, z_n \in \mathcal{O}$ denote the standard coordinates on $\mathbb{C}^n$ restricted to $M$ and use multi-index notation $\mu=(\mu_1,\ldots,\mu_n)\in\mathbb{N}^n$ with $z^\mu=z_1^{\mu_1}\cdots z_n^{\mu_n}$, $\mu!=\mu_1!\cdots\mu_n!$ and $|\mu|=\mu_1+\ldots+\mu_n$. It's easy to show that an operator of the form $$ \sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}: \mathcal{O} \to \mathcal{O}$$ with $c_\mu\in \mathcal{O}$, satisfies the algebraic definition of a $\mathbb{C}$-linear differential operator of rank at most $k$.

To show the converse, assume $D\colon \mathcal{O} \to \mathcal{O}$ is such an algebraic DO of rank at most $k$. We want to show $D = \sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}$ for suitable coefficients $c_\mu \in \mathcal{O}$. Define them recursively as

$$ c_{\mu}= \cases{ D (1) \quad \text{ when } \mu=0 \\ \left(D - \sum\limits_{0\leq |\nu|< |\mu|} c_\nu \frac{\partial^{|\nu|}}{\partial z^\nu}\right)\left(\frac{z^\mu}{\mu!}\right) \quad \text{ when } 1\leq |\mu|\leq k. } $$ Its straightforward to checkt that $D$ and $\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}$ agree on polynomials in $z$ of degree at most $k$.

To show that they also agree on any other $u\in \mathcal{O}$, let $x\in M$ and use Taylor expansion around $x$ to write $u=p+r$ where $p$ is a polynomial in $z$ of degree at most $k$ and $r$ is an element of $I^{k+1}_x\subset \mathcal{O}$, the $k+1$st power of the vanishing ideal at $x$. Now use the fact that $D(I^{k+m})=I^m$, for any ideal $I\subset \mathcal{O}$ and any algebraic DO $D$ of rank at most $k$. So that $$ (D u)(x)=D(p+r)(x)=(D p)(x)= \left(\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}p\right)(x)=\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}(p+r)(x)=\left(\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu} u\right)(x). $$ Since this holds for any $x$ in $M$ we conclude $D u =\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu} u$.

Q1 should now be easy: since the topology of $M$ as complex manifold agrees with its topology as real manifold, by Peetres theorem a $\mathbb{C}$-linear local operator $D : C^\infty(M,\mathbb{C}) \to C^\infty(M,\mathbb{C})$ satisfies the algebraic definition of a DO with smooth $f$ (in the sense of your post $fD - Df$ being of lower rank). Since holomorphic $f$ are also smooth, the algebraic definition holds also for holomorphic $f$. Now if $D$ moreover preserves the subring of holomorphic functions it is a complex differential operator by Q2.

Yes, for both the second question.

Q2: Since it's a local question, it suffices to consider $M$ open in $\mathbb{C}^n$ and $\mathcal{E},\mathcal{F}$ one dimensional trivial complex bundles. So let $\mathcal{O}$ denote the ring of holomorphic functions on $M$, let $z_1,\ldots, z_n \in \mathcal{O}$ denote the standard coordinates on $\mathbb{C}^n$ restricted to $M$ and use multi-index notation $\mu=(\mu_1,\ldots,\mu_n)\in\mathbb{N}^n$ with $z^\mu=z_1^{\mu_1}\cdots z_n^{\mu_n}$, $\mu!=\mu_1!\cdots\mu_n!$ and $|\mu|=\mu_1+\ldots+\mu_n$. It's easy to show that an operator of the form $$ \sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}: \mathcal{O} \to \mathcal{O}$$ with $c_\mu\in \mathcal{O}$, satisfies the algebraic definition of a $\mathbb{C}$-linear differential operator of rank at most $k$.

To show the converse, assume $D\colon \mathcal{O} \to \mathcal{O}$ is such an algebraic DO of rank at most $k$. We want to show $D = \sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}$ for suitable coefficients $c_\mu \in \mathcal{O}$. Define them recursively as

$$ c_{\mu}= \cases{ D (1) \quad \text{ when } \mu=0 \\ \left(D - \sum\limits_{0\leq |\nu|< |\mu|} c_\nu \frac{\partial^{|\nu|}}{\partial z^\nu}\right)\left(\frac{z^\mu}{\mu!}\right) \quad \text{ when } 1\leq |\mu|\leq k. } $$ Its straightforward to checkt that $D$ and $\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}$ agree on polynomials in $z$ of degree at most $k$.

To show that they also agree on any other $u\in \mathcal{O}$, let $x\in M$ and use Taylor expansion around $x$ to write $u=p+r$ where $p$ is a polynomial in $z$ of degree at most $k$ and $r$ is an element of $I^{k+1}_x\subset \mathcal{O}$, the $k+1$st power of the vanishing ideal at $x$. Now use the fact that $D(I^{k+m})=I^m$, for any ideal $I\subset \mathcal{O}$ and any algebraic DO $D$ of rank at most $k$. So that $$ (D u)(x)=D(p+r)(x)=(D p)(x)= \left(\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}p\right)(x)=\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu}(p+r)(x)=\left(\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu} u\right)(x). $$ Since this holds for any $x$ in $M$ we conclude $D u =\sum\limits_{|\mu|\leq k}c_\mu \frac{\partial^{|\mu|}}{\partial z^\mu} u$.

Edit: After the clarification of the question what follows is not a correct answer.

Q1 should now be easy: since the topology of $M$ as complex manifold agrees with its topology as real manifold, by Peetres theorem a $\mathbb{C}$-linear local operator $D : C^\infty(M,\mathbb{C}) \to C^\infty(M,\mathbb{C})$ satisfies the algebraic definition of a DO with smooth $f$ (in the sense of your post $fD - Df$ being of lower rank). Since holomorphic $f$ are also smooth, the algebraic definition holds also for holomorphic $f$. Now if $D$ moreover preserves the subring of holomorphic functions it is a complex differential operator by Q2.