Suppose that $V, W$ are vector spaces and $f, g : V \to W$ are two parallel morphisms such that $f \le g$, for some preorder $\le$ satisfying your conditions.

Claim: $f$ is a scalar multiple of $g$.

Proof. The first condition implies that if $v : 1 \to V$ is any vector in $V$ (here $1$ denotes the $1$-dimensional vector space) then $f \circ v \le g \circ v$. The second condition implies that there is a monomorphism from the image of $f \circ v$ to the image of $g \circ v$. This is equivalent to the condition that for all $v \in V$ there is a scalar $\lambda(v)$, possibly zero, such that

$$f(v) = \lambda(v) g(v).$$

Observe that

$$f(v + v') = \lambda(v + v') (g(v) + g(v')) = f(v) + f(v') = \lambda(v) g(v) + \lambda(v') g(v').$$

Hence if $g(v)$ and $g(v')$ are linearly independent it follows that $\lambda(v) = \lambda(v') = \lambda(v + v')$. Now we split into cases.

Case: $\text{im}(g)$ has dimension at least $2$. Then the equivalence relation on vectors in $\text{im}(g)$ generated by "linearly independent" identifies every vector, and it follows that $\lambda(v)$ is a constant $\lambda$ and $f = \lambda g$.

Case: $\text{im}(g)$ has dimension $1$. Then $\text{im}(f)$ is contained in $\text{im}(g)$, and hence has dimension at most $1$. Picking $v_0 \in V$ such that $g(v_0) \neq 0$, we then have $f(v_0) = \lambda(v_0) g(v_0)$, and hence it again follows that $f = \lambda g$ (where $\lambda = \lambda(v_0)$).

Case: $\text{im}(g)$ has dimension $0$. Then $f = g = 0$.

This concludes the proof. $\Box$

So possibilities are limited. Over any field, we can take $f \le g$ iff $f = \lambda g$ for some $\lambda$. This is a preorder, but it's pretty boring: linear maps which are nonzero scalar multiples of each other are isomorphic, and the only order relation which isn't an isomorphism is that $0 \le f$ for any nonzero $f$. Over $\mathbb{R}$ we can take $f \le g$ iff $f = \lambda g$ for some $0 \le \lambda \le 1$. This is even a partial order, but it's still not very interesting.

Suppose that $V, W$ are vector spaces and $f, g : V \to W$ are two parallel morphisms such that $f \le g$, for some preorder $\le$ satisfying your conditions.

Claim: $f$ is a scalar multiple of $g$.

Proof. The first condition implies that if $v : 1 \to V$ is any vector in $V$ (here $1$ denotes the $1$-dimensional vector space) then $f \circ v \le g \circ v$. The second condition implies that there is a monomorphism from the image of $f \circ v$ to the image of $g \circ v$ (Edit: here I interpret the condition to mean that the monomorphism must be compatible with the inclusion into $W$, but as Eric Wofsey's answer shows, this extra assumption is unnecessary). This is equivalent to the condition that for all $v \in V$ there is a scalar $\lambda(v)$, possibly zero, such that

$$f(v) = \lambda(v) g(v).$$

Observe that

$$f(v + v') = \lambda(v + v') (g(v) + g(v')) = f(v) + f(v') = \lambda(v) g(v) + \lambda(v') g(v').$$

Hence if $g(v)$ and $g(v')$ are linearly independent it follows that $\lambda(v) = \lambda(v') = \lambda(v + v')$. Now we split into cases.

Case: $\text{im}(g)$ has dimension at least $2$. Then the equivalence relation on vectors in $\text{im}(g)$ generated by "linearly independent" identifies every vector, and it follows that $\lambda(v)$ is a constant $\lambda$ and $f = \lambda g$.

Case: $\text{im}(g)$ has dimension $1$. Then $\text{im}(f)$ is contained in $\text{im}(g)$, and hence has dimension at most $1$. Picking $v_0 \in V$ such that $g(v_0) \neq 0$, we then have $f(v_0) = \lambda(v_0) g(v_0)$, and hence it again follows that $f = \lambda g$ (where $\lambda = \lambda(v_0)$).

Case: $\text{im}(g)$ has dimension $0$. Then $f = g = 0$.

This concludes the proof. $\Box$

So possibilities are limited. Over any field, we can take $f \le g$ iff $f = \lambda g$ for some $\lambda$. This is a preorder, but it's pretty boring: linear maps which are nonzero scalar multiples of each other are isomorphic, and the only order relation which isn't an isomorphism is that $0 \le f$ for any nonzero $f$. Over $\mathbb{R}$ we can take $f \le g$ iff $f = \lambda g$ for some $0 \le \lambda \le 1$. This is even a partial order, but it's still not very interesting.

Suppose that $V, W$ are vector spaces and $f, g : V \to W$ are two parallel morphisms such that $f \le g$, for some preorder $\le$ satisfying your conditions.

Claim: $f$ is a scalar multiple of $g$.

Proof. The first condition implies that if $v : 1 \to V$ is any vector in $V$ (here $1$ denotes the $1$-dimensional vector space) then $f \circ v \le g \circ v$. The second condition implies that there is a monomorphism from the image of $f \circ v$ to the image of $g \circ v$. This is equivalent to the condition that for all $v \in V$ there is a scalar $\lambda(v)$, possibly zero, such that

$$f(v) = \lambda(v) g(v).$$

Observe that

$$f(v + v') = \lambda(v + v') (g(v) + g(v')) = f(v) + f(v') = \lambda(v) g(v) + \lambda(v') g(v').$$

Hence if $g(v)$ and $g(v')$ are linearly independent it follows that $\lambda(v) = \lambda(v') = \lambda(v + v')$. Now we split into cases.

Case: $\text{im}(g)$ has dimension at least $2$. Then the equivalence relation on vectors in $\text{im}(g)$ generated by "linearly independent" identifies every vector, and it follows that $\lambda(v)$ is a constant $\lambda$ and $f = \lambda g$.

Case: $\text{im}(g)$ has dimension $1$. Then $\text{im}(f)$ is contained in $\text{im}(g)$, and hence has dimension at most $1$. Picking $v_0 \in V$ such that $g(v_0) \neq 0$, we then have $f(v_0) = \lambda(v_0) g(v_0)$, and hence it again follows that $f = \lambda g$ (where $\lambda = \lambda(v_0)$).

Case: $\text{im}(g)$ has dimension $0$. Then $f = g = 0$.

This concludes the proof. $\Box$

So possibilities are limited. Over any field, "$f \le g$ iff $f = \lambda g$" defines a preorder, but it's pretty boring: linear maps which are nonzero scalar multiples of each other are isomorphic, and the only order relation which isn't an isomorphism is that $0 \le f$ for any nonzero $f$. Over $\mathbb{R}$, or more generally any field whose group of units contains an ordered subgroup, we can take "$f \le g$ iff $f = \lambda g$ where $0 \le \lambda \le 1$." This is even a partial order, but it's still not very interesting.

Suppose that $V, W$ are vector spaces and $f, g : V \to W$ are two parallel morphisms such that $f \le g$, for some preorder $\le$ satisfying your conditions.

Claim: $f$ is a scalar multiple of $g$.

Proof. The first condition implies that if $v : 1 \to V$ is any vector in $V$ (here $1$ denotes the $1$-dimensional vector space) then $f \circ v \le g \circ v$. The second condition implies that there is a monomorphism from the image of $f \circ v$ to the image of $g \circ v$. This is equivalent to the condition that for all $v \in V$ there is a scalar $\lambda(v)$, possibly zero, such that

$$f(v) = \lambda(v) g(v).$$

Observe that

$$f(v + v') = \lambda(v + v') (g(v) + g(v')) = f(v) + f(v') = \lambda(v) g(v) + \lambda(v') g(v').$$

Hence if $g(v)$ and $g(v')$ are linearly independent it follows that $\lambda(v) = \lambda(v') = \lambda(v + v')$. Now we split into cases.

Case: $\text{im}(g)$ has dimension at least $2$. Then the equivalence relation on vectors in $\text{im}(g)$ generated by "linearly independent" identifies every vector, and it follows that $\lambda(v)$ is a constant $\lambda$ and $f = \lambda g$.

Case: $\text{im}(g)$ has dimension $1$. Then $\text{im}(f)$ is contained in $\text{im}(g)$, and hence has dimension at most $1$. Picking $v_0 \in V$ such that $g(v_0) \neq 0$, we then have $f(v_0) = \lambda(v_0) g(v_0)$, and hence it again follows that $f = \lambda g$ (where $\lambda = \lambda(v_0)$).

Case: $\text{im}(g)$ has dimension $0$. Then $f = g = 0$.

This concludes the proof. $\Box$

So possibilities are limited. Over any field, we can take $f \le g$ iff $f = \lambda g$ for some $\lambda$. This is a preorder, but it's pretty boring: linear maps which are nonzero scalar multiples of each other are isomorphic, and the only order relation which isn't an isomorphism is that $0 \le f$ for any nonzero $f$. Over $\mathbb{R}$ we can take $f \le g$ iff $f = \lambda g$ for some $0 \le \lambda \le 1$. This is even a partial order, but it's still not very interesting.