I have the linear transforation $f: W \to C^\infty(\mathbb{R})$ given by $f(x(t)=x''(t)-x'(t)$
$W$ is $span\{t,e^t,e^{-t}\}$ and the function $\frac{1}{2}e^{-t}+t \in W$
I found the image of f to be $Im(f)=e^{-t}-1$
"What is the dimension of the image f(W)? Determine the kernel for f"
I know that $dim(W)=dim(ker(f))+dim(f(W))$, but I'm confused as to what the dimension of the image is? What meaning does it have to the dimension of an image of a linear transformation if there's a constant in it?
First of all your notation $Im(f)=e^{-t}-1$ is not valid because on LHS you have a set of function while the RHS is a function...
We see that $\{t, e^t, e^{-t}\}$ is a linearly independent set of functions, so we get $dim(W)=3$. (I hope this is clear)
Since $f$ is a linear map, it is sufficient to consider $$ f(t)=-1,~f(e^t)=0,~f(e^{-t})=e^{-t}-(-e^{-t})=2e^{-t} $$ because each linear map is determined by the way how it maps the base. You can compute from it the image and the kernel.
So your image is given by \begin{align} f(W)&=\{f(at+be^{t}+ce^{-t})~:~a,b,c\in\mathbb R\} =\{af(t)+bf(e^t)+cf(e^{-t})~:~a,b,c\in\mathbb R\} \\&=\{a(-1)+2ce^{-t}~:~c\in\mathbb R\}=span\{-1,e^{-1}\} \end{align} Since $\{-1,e^{-t}\}$ is independent, you get $\dim(f(W))=2$.
You can also see directly the kernel of $f$, but you should do it yourself.