Let $X=Y_1\sqcup Y_2$, with both $Y_i$ homeomorphic to $Y$. Then $C(X)=C(Y_1)\oplus C(Y_2)$. Given $a\in C(Y)$ let $\phi\colon a\mapsto a\oplus 0$, in the obvious way. This $\phi$ cannot be any $f^*$, since $f^*$ would necessarily map $1\mapsto 1\oplus 1$. I believe, this is a counter example to your putative theorem, which shows that you may want a connectedness hypothesis on your spaces.
For more general information, I second Ramsey's recommendation of "Rings of continuous functions" by Gillman and Jerison. Though, I don't think it has the exact theorem you are looking for.
The strongest relevant result from that book is Theorem 10.8, which states that a homomorphism $\mathfrak{s}\colon C(Y)\to C(X)$ determines a unique continuous map $\tau\colon E\to \upsilon Y$ with the properties like what you want. Here $E$ is a clopen subset of $X$ and $\upsilon Y$ is the (Hewitt) realcompactification of $Y$, which is a bigger space than $Y$. See the book for full details. The hypotheses on $X$ and $Y$ (implicitly) include complete regularity, which is weaker than local compactness. Note that the homomorphism $\mathfrak{s}$ is not assumed to be continuous in any topologies on $C(Y)$ and $C(X)$. Perhaps your continuity requirement is enough to cut $\upsilon Y$ down to $Y$ and give you the desired result.
If you mean by LF-space a strict inductive limit of Frechet spaces (as it was done by Dieudonne and Schwartz) I think the answer is yes.
Here is what I believe could be made a proof: Since the inductive tensor product
respects inductive limits you have $C^\infty(M) \otimes_i F = \lim C^\infty(M) \otimes_i F_n$ if $F= \lim F_n$ is a strict LF-space. Since $F_n$ is a closed topological
subspace of $F$ you get that $C^\infty(M) \otimes_i F_n = C^\infty(M) \otimes_\pi F_n$
and this implies that $C^\infty(M) \otimes_i F = \lim C^\infty(M) \otimes_\pi F_n$ is a dense topological subspace of $C^\infty(M) \otimes_\pi F$. On the other hand, it is complete
because it remains strict (as for as I remember this is due either to Dieudnne-Schwartz or to Köthe).
EDIT: Sorry, I confused the Frechet case with the DF case -- only in the latter case
$\lim E \otimes_\pi F_n$ is a topological subspace of $E\otimes_\pi \lim F_n$.
For strict LF-spaces the answer to your question was already given in Grothendieck's thesis (part I, page 47): If E is a proper (non-normable) Frechet space and $F=\lim F_n$ is a strict inductive limit with a strictly increasing sequence $F_n$ then
$\lim E\otimes_\pi F_n$ is NEVER a topological subspace of $E \otimes_\pi \lim F_n$.
On the other hand, if the nuclear Frechet space $E$ has a continuous norm (as it is the case for $C^\infty(M)$) and $F=\lim F_n$ is a strict LF-space then
$\lim E\otimes_\pi F_n = E\otimes_\pi F$ holds algebraically:
The latter space is the space of all continuous linear operators $T$ from $E'$ to $F$ and the former of those operators with values in some $F_n$. But $E'$ contains a bounded set
(the polar of the $0$-neighborhood corresponding to the continuous norm) which is total, i.e., its linear span is dense in $E'$. For any continuous $T$ the bounded set $T(B)$ is contained in some $F_n$ (by the so called regularity of strict LF-spaces) and since $F_n$ is closed in $F$ one gets from the continuity $T(E') \subseteq \overline{[T(B)]} \subseteq F_n$.
There might be non-strict inductive limits where $\lim E \otimes_\pi F_n = E\otimes_\pi\lim F_n$ holds topologically but I do not know an example since both, the algebraic and the topological, conditions are very restrictive. If $F$ is the dual of a nuclear Frechet space $X$ then the arguments above should show that the algebraic coincidence is equivalent to
$L(X,E) = LB(X,E)$, the space of all operators from mapping some $0$-neighborhood of $X$ into a bounded subset of $E$. This situation has been investigated by Vogt,
J. Reine Angew. Math. 345 (1983), 182–200, and it implies Ext$^1(E,X)=0$. On the other
hand, I believe results of Grothendieck and Vogt yield that the topological coincidence
implies Ext$^1(X,E)=0$. If $X$ and $E$ are both power series spaces these conditions contradict each other.
In conclusion: The topological equality $C^\infty(M) \otimes_i F = C^\infty(M) \otimes_\pi F$ seems very unlikely whenever $F$ is not a Frechet space.
Best Answer
Hi.
One issue with Bochner integration is that it does not include Riemann-integration. There are Banach-space-valued R-integrable functions that are not B-integrable (example: Consider $X:=\mathcal{l}^p([0,1])$ with $2\leq p < \infty$ and $f:[0,1]\to X, f(t):=e_t$ where $e_t$ is the tupel with exactly one equal to 1 and all other components equal to 0.). The Gelfand-Pettis-Integral on the other hand includes both the Bochner- and the Riemann-Integral.
One problem with B-integration is that you need functions that are almost separable valued (meaning: there is a nullset whose complement has separable image) in order to approximate them with simple functions and this may be a strong restriction. Another issue is that for certain applications the weak topologies just behave better than the strong ones so that Pettis-integration is the natural notion of integration in this cases.