I thought a little bit about your question, which is phrased a little more generally than I like, but I decided to think about it with the restrictions that $\widehat{\Delta^0}$ be a differential operator and $\widehat{d^\ast}$ be $d^\ast$ plus a lower-order (i.e., zeroth order) differential operator. I also decided to think, not of the most general perturbation of $\widehat{\Delta^1}$, but of perturbations of the form $\widehat{\Delta^1} = \nabla^\ast\nabla+L$, where $L$ means scalar multiplication by a function $L$ on $M$.
The first thing to notice is that, since the principal symbol $\sigma^1_\xi$ of $\widehat{\Delta^1}$ is just scalar multiplication by $|\xi|^2$ and the principal symbol of $\widehat{d^\ast}$ is $\alpha\mapsto \xi\cdot\alpha$, it follows from the symbol calculus that the principal symbol of $\widehat{\Delta^0}$ must also be scalar multiplication by $|\xi|^2$, i.e., the differential operator $\widehat{\Delta^0} - \Delta^0$ must be of order at most $1$. However, it must also be self-adjoint, and this implies that it must be of order $0$, i.e., that $\widehat{\Delta^0} = \Delta^0 + H$ (where '$H$' means scalar multiplication by a smooth function $H$ on $M$).
Now, $\widehat{d^\ast}\alpha = d^\ast\alpha + \phi\cdot\alpha$ for some $1$-form $\phi$ on $M$. Under the given assumptions, it is not hard to see that the equation
$$
\widehat{d^\ast}\widehat{\Delta^1} = \widehat{\Delta^0}\widehat{d^\ast}
$$
implies that $\widehat{\Delta^0} = \Delta^0 + H$ for some function $H$ on $M$.
(This is what I just explained above.)
Now, the operator $E = \widehat{d^\ast}\widehat{\Delta^1} - \widehat{\Delta^0}\widehat{d^\ast}$, when expanded out, is of first order (not second order, as you might have expected). Looking at the principal symbol of $E$ and setting this equal to zero gives $4$ equations, and these are equivalent to $\nabla\phi = f\ g$, where, for simplicity, I have set $f = \tfrac12(L-K-H)$. Taking the covariant derivative of both sides of $\nabla\phi = f\ g$ and using the definition of $K$, one gets $df = -K\ \phi$. Substituting this back into the expression for $E$, it reduces to a $0$-th order operator which turns out to be $E(\alpha) = (2f\phi +dK - dL)\cdot\alpha$. Setting this equal to zero gives $d(|\phi|^2 + K - L) = 0$, since $d\bigl(|\phi|^2\bigr) = 2f\phi$ (which is a consequence of $\nabla\phi = f\ g$). Thus, $L = K + |\phi|^2 - c$ for some constant $c$.
Thus, one finds that one must have the identities
$$
\nabla \phi = f\ g\qquad\text{and}\qquad df = -K\ \phi
$$
for some function $f$ on $M$ while
$$
L = K + |\phi|^2 - c\qquad\text{and}\qquad H = |\phi|^2 -2f - c
$$
for some constant $c$.
Obviously, there is always the trivial solution $(\phi,f) = (0,0)$, which gives the well-known intertwining of the Hodge Laplacian on $1$-forms and $0$-forms. More interesting is the case when there are nontrivial solutions $(\phi,f)$ to the above equations. This puts severe restrictions on the metric $g$, but these can be understood.
Let's assume that $M$ is connected and complete. Then the pair of equations $\nabla\phi = f\ g$ and $df = -K\ \phi$ forms a linear total differential system, so if the pair $(\phi,f)$ vanishes anywhere, it vanishes identically. Let's assume that it does not. The equation $\nabla\phi = f\ g$ implies that $d\phi = 0$ and that the vector field $F$ that is $g$-dual to $\ast\phi$ is a Killing field. Thus, $(M,g)$ is, at least locally, a surface of revolution. Any fixed points of $F$ are isolated elliptic points. Write $\phi = du$ for some function $u$ (at the moment locally defined). Note that the critical points of $u$ are nondegenerate and of index 0 or 2. It is then not hard to show that $|\phi|^2 = a(u)$ for some function $a$ and that, in the region where $a>0$, the metric $g$ takes the form
$$
g = \frac{du^2}{a(u)} + a(u)\ d\psi^2
$$
for some (locally defined) function $\psi$. In these coordinates, one has $\phi = du$, $f = \tfrac12a'(u)$, and $K = -\tfrac12a''(u)$. One also has $L = a(u) -\tfrac12a''(u)- c$ and $H = a(u) - a'(u) - c$.
Conversely, if one starts with a function $a$ positive on some domain on the $u$-line, then the above formulae give a solution to the intertwining equation. If $a$ is positive and periodic on the $u$-line, then one obtains a solution on the torus.
One can also obtain solutions on the $2$-sphere: If $a$ is smooth and satisfies $a(u_0) = a(u_1) = 0$ (where $ u_0 < u_1 $) while $a$ is positive between $u_0$ and $u_1$ and satisfies $a'(u_0) = -a'(u_1) = b >0$, then the metric
$$
g = \frac{du^2}{a(u)} + \frac{4a(u)}{b^2}\ d\theta^2
$$
defines a smooth metric on the $2$-sphere with $(u,\theta)$ as 'polar coordinates' (with $\theta$ being $2\pi$-periodic and $u_0\le u\le u_1$, with $u=u_i$ corresponding to the 'poles' of rotation), and this gives a family of solutions to the intertwining equation.
To get $\widehat{\Delta^1}$ to be the Bochner Laplacian, one must have $L = 0$, which is equivalent to $a''(u) = 2\bigl(a(u)-c\bigr)$. This has 'spherical solutions', such as
$$
a(u) = c - e\ \cosh\bigl(\sqrt{2}\ u\bigr),
$$
where the constants $c$ and $e$ satisfy $c > e > 0$. Thus, there is a nontrivial $2$-parameter family of metrics on the $2$-sphere such that the Bochner Laplacian has the desired intertwining property.
A similar calculation can be done for the more general case when $\widehat{d^\ast}$ is allowed to be somewhat more general (but still underdetermined elliptic), but I won't go into that unless someone is interested. I'll just say that, aside from the metrics above, the only metrics that admit a nontrivial solution when $\widehat{d^\ast}$ is an underdetermined elliptic first order differential operator are the metrics of the form
$$
g = \bigl( a\ (x^2{+}y^2) + 2b\ x + c\bigr)\ \bigl(dx^2 + dy^2\bigr)
$$
where $a$, $b$, and $c$ are constants such that the open set $M$ in the $xy$-plane defined by $a\ (x^2{+}y^2) + 2b\ x + c > 0$ is nonempty. Thus, the metrics that allow this are very restricted.
Best Answer
Ad 1: Yes, there is. The formula is $$\nabla^*(X^\flat \otimes u) = - \nabla_X u -\mathrm{div}(X) \cdot u,$$ as can easily seen by local computation. Here, $X$ is a vector field and $X^\flat$ is the dual one form w.r.t. the metric.
Note that unless you have a scalar product on the bundle $E$, too, the dual operator will be an operator $\Gamma^\infty(T^*M \otimes E^*) \longrightarrow \Gamma^\infty(E^*)$. The above formula holds in both cases (depending on which case you are in, $u$ will be a section of $E$ or $E^*$, and the connection on the right will be either your given connection on $E$ or the corresponding dual connection on $E^*$ (this connection is always well-defined by the formula $(\nabla u)(s) = d(u(s)) - u(\nabla(s))$).
A similar formula holds even in the case that you don't have a metric on $M$, in which case the dual operator will send sections of $TM \otimes E^* \otimes |\Lambda|$ to sections of $E^* \otimes |\Lambda|$, where $|\Lambda|$ is the density bundle.
Ad 2: Let me first say that in my opinion, one uses the word "formal" only to indicate that one doesn't bother about any functional analytic meaning of the word adjoint whatsoever, and one does not automatically want to say that there has to be any setting where there is an "actual" adjoint.
However, to give a more constructive answer: Yes, this operator is always an adjoint operator in the functional analytic sense.
First, this is true for elliptic differential operators on manifolds. On compact manifolds, they have a unique closed extension, whose adjoint operator is given on smooth functions by the formal adjoint.
In general, consider for example the space $\mathscr{D}=\mathscr{D}(M, E)$ of test sections (which is not a Fréchet space unless $M$ is compact!). It can be embedded into its dual $\mathscr{D^\prime}(M, E^*)$. Now any differential operator $A$ is a continuous operator on $\mathscr{D}$, and its adjoint $A^\prime$ is an operator on $\mathscr{D}^\prime$. However, on the elements $\mathscr{D}\subset \mathscr{D}^\prime$ $A^\prime$ acts just as the formal adjoint of $A$. In this sense, the formal adjoint is also an "actual" adjoint in some sense.
However, this is quite formal and not at all deep. It follows at once from the definitions of distributions etc.